PAT从零单排

时间：2015-01-16 23:36:53 阅读：329 评论：0 收藏：0 [点我收藏+]

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偶然原因开始了PAT之旅
前前后后用了3天时间刷完了PAT Basic（35题）
现在记录下从零单排的过程，总的来说题目很简单
当然有个别题目的坑点还是比较给力的，准确的说，题目的数据很给力

1001. 害死人不偿命的(3n+1)猜想 (15)

纯模拟题，是偶数除以2，奇数（3n+1）除以2，到1结束

#include <cstdio> 
#include <cstring>
#include <algorithm>
using namespace std;

int n, step;

void scan( int& x )
{
    char c;
    while( c = getchar(), c < ‘0‘ || c > ‘9‘ );
    x = c - ‘0‘;
    while( c = getchar(), c >= ‘0‘ && c <= ‘9‘ ) x = x * 10 + c - ‘0‘;
}

int main()
{
    scan( n );
    step = 0;
    while( n != 1 )
    {
        if( n & 1 )
            n = ( n * 3 + 1 ) >> 1;
        else
            n >>= 1;
        step++;
    }
    printf( "%d\n", step );
    return 0;
}

代码君

1002. 写出这个数 (20)

把一个数串各个数位上的数加起来，然后用给定格式输出

#include <cstdio> 
#include <cstring>
#include <algorithm>
using namespace std;

char str[110];
int i, j;
int res, flag;
char du[15][5] = {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};

void out( int x )
{
    if( x > 9 )
        out( x/10 );
    if( flag ) putchar( ‘ ‘ );
    else flag = 1;
    printf( "%s", du[ x % 10 ] );
}

int main()
{
    gets(str);
    res = flag = 0;
    for( i = 0; str[i]; ++i )
        res += str[i] - ‘0‘;
    out( res );
    return 0;
}

代码君

1003. 我要通过！(20)

xPyTz 满足X*Y=Z 的PAT串即为正确答案，否则为错误的串
如 AAPAAPAAAA 为正确的串

#include <cstdio>
#include <cstring>
using namespace std;

int n;
char s[ 110 ];

bool check( char* str )
{
    int len = strlen( str );
    if( len < 3 ) return false;
    
    int cl = 0, cm = 0, cr = 0, cnt = 0;
    int fa = 0, fp = 0, ft = 0;
    
    for( int i = 0; i < len; ++i )
    {
        if( str[i] == ‘A‘ )    fa++;
        else if( str[i] == ‘P‘ ) fp++;
        else if( str[i] == ‘T‘ ) ft++;
        else    return false;
    }
    if( fp != 1 || ft != 1 )
        return false;
    while( str[ cnt++ ] != ‘P‘ )
    {
        cl++;
        if( cnt >= len )
            return false;
    }
    while( str[ cnt++ ] != ‘T‘ )
    {
        cm++;
        if( cnt >= len )
            return false;
    }
    while( str[ cnt++ ] == ‘A‘ )
    {
        cr++;
        if( cnt == len )
            break;
    }
    if( cl * cm == cr )
        return true;
    else
        return false;
}


int main()
{
    scanf( "%d", &n );
    while( n-- )
    {
        scanf( "%s", s );
        if( check( s )) 
            puts( "YES" );
        else
            puts( "NO" );
    } 
    return 0;
}

代码君

1004. 成绩排名 (20)

简单排序

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct score
{
    char name[15];
    char id[15];
    int score;
}S[1000];

int cmp( score a, score b )
{
    if( a.score == b.score )
        return strcmp( a.id , b.id );
    return a.score > b.score;
}

int n;

int main()
{
    scanf( "%d", &n );
    for( int i = 0; i < n; ++i )
        scanf( "%s %s %d", &S[i].name, &S[i].id, &S[i].score );
    sort( S, S + n, cmp );
    printf( "%s %s\n%s %s\n", S[0].name, S[0].id, S[n-1].name, S[n-1].id );
    return 0;
}

代码君

1005. 继续(3n+1)猜想 (25)

Hash标记一下，over

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int H[1000000], n, x, flag, num[200] ;

int main()
{
    memset( H, 0 , sizeof( H ) );
    scanf( "%d", &n );
    for( int i = 0; i < n; ++i )
    {
        scanf( "%d", &num[i] );
        x = num[i];
        while( x != 1 )
        {
            if( x & 1 )
                x = ( x * 3 + 1 ) >> 1;
            else
                x >>= 1;
            H[ x ] = 1;
        }
    }
    sort( num, num + n );
    for( int i = n-1; i >= 0; --i )
    {
        if( !H[ num[i] ] )
        {
            if( flag )    putchar( ‘ ‘ );
            else flag = 1; 
            printf( "%d", num[i] );
        }
    }
    putchar( ‘\n‘ );
    return 0;
}

代码君

1006. 换个格式输出整数 (15)

题目要换格式你就照着换格式就好了

#include <cstdio>
#include <cstring>
using namespace std;

int n, x, y, z;
void doit()
{
    scanf( "%d", &n );
    x = n / 100, y = n % 100 /10 , z = n % 10;
    if( x )    
        while( x-- )    putchar( ‘B‘ );
    if( y )
        while( y-- )    putchar( ‘S‘ );
    if( z )
    {
        for( int i = 1; i <= z ; ++i )
            printf( "%d", i );
    }
    putchar( ‘\n‘ );
}

int main()
{
    doit();    
    return 0;
}

代码君

1007. 素数对猜想 (20)

打表+递推

#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define MAXN 100000

int vis[ MAXN + 10 ], prime[ MAXN + 10 ], n;
int d[ MAXN + 10], H[ MAXN + 10 ], res[ MAXN + 10 ], cnt;

void init()
{
    int m = sqrt( MAXN + 0.5 );
    for( int i = 2; i <= m; ++i ) if( !vis[ i ] )
        for( int j = i * i; j <= MAXN; j += i )    vis[ j ] = 1;
    cnt = 0;
    for( int i = 2; i <= MAXN; ++i )
    {
        if( !vis[i] )
        {
            prime[ cnt ] = i;
            H[ i ] = cnt;
            cnt++;
        }
    }
        
    for( int i = 1; i < cnt; ++i )
    {
        if( prime[i] - prime[i-1] == 2 )
            d[ i ] = 1;
    }
    for( int i = 1; i < cnt; ++i )
    {
        res[i] = res[i-1] + d[i];
    }
}

int main()
{
    init();
    scanf( "%d", &n );
    while( vis[n] )    n--;
    printf( "%d\n", res[H[n]] );

    return 0;
}

代码君

1008. 数组元素循环右移问题 (20)

简单水题，连接一下然后取模就是了

#include <cstdio>
#include <cstring>
using namespace std;

int x[222], n, m;

int main()
{
    scanf( "%d %d", &n, &m );
    for( int i = 0; i < n; ++i )
    {
        scanf( "%d", &x[i] );
        x[i+n] = x[i];        
    }
    int s = m % n;
    for( int i = n - s, j = 0; j < n; ++i, ++j )
    {
        if( j != 0 )    putchar( ‘ ‘ );
        printf( "%d", x[i] );
    }
    return 0;
}

代码君

1009. 说反话 (20)

那就说一次反话咯

#include <cstdio>
#include <cstring>
using namespace std;

char str[88][88];

int main()
{
    int ct = 0;
    while( ~scanf( "%s", str[ct++] ) );
    printf( "%s", str[ ct - 2 ] );
    for( int i = ct - 3; i >= 0; --i )
    {
        printf( " %s", str[i] );
    }
    putchar( ‘\n‘ );
    return 0;
}

代码君

1010. 一元多项式求导 (25)

简单求导，注意特判

#include <cstdio>
#include <cstring>
using namespace std;

int s[1010][2], cnt = 0, x, y;

int main()
{
    while( ~scanf( "%d %d", &x, &y ) && y )
    {
        s[cnt][0] = x;
        s[cnt++][1] = y;
    }
    if( cnt == 0 )
        puts( "0 0" );
    else
    {
        for( int i = 0; i < cnt; ++i )
        {
            s[i][0] *= s[i][1];
            s[i][1]--;
            if( i == 0 )
                printf( "%d %d", s[i][0], s[i][1] );
            else
                printf( " %d %d", s[i][0], s[i][1] );
        }
        putchar( ‘\n‘ );        
    }
    
    return 0;
}

代码君

1011. A+B和C (15)

水题。直接用long long 免除后顾之忧

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;

LL A, B, C;
int n;

int main()
{
    scanf( "%d", &n );
    for( int i = 1; i <= n; ++i )
    {
        scanf( "%lld %lld %lld", &A, &B, &C );
        printf( "Case #%d: ", i );
        if( A + B > C )    puts( "true" );
        else            puts( "false" );
    }
    return 0;
}

代码君

1012. 数字分类 (20)

按题意分类，有个double注意一下

#include <cstdio>
#include <cstring>
using namespace std;

int x[1010], n;
int res[5], flag[5];
double sum;
int t = 1;

void check( int key )
{
    if( key % 5 == 0 && key % 2 == 0 ) res[0] += key, flag[0] = 1;
    if( key % 5 == 1 )     { res[1] = res[1] + key * t; t = -t; flag[1] = 1; }
    if( key % 5 == 2 )    res[2]++, flag[2] = 1;
    if( key % 5 == 3 )    { res[3]++; sum += key; flag[3] = 1; }
    if( key % 5 == 4 )    if( key > res[4] ) res[4] = key, flag[4] = 1;
}

int main()
{
    scanf( "%d", &n );
    for( int i = 0; i < n; ++i )
    {
        scanf( "%d", &x[i] );
        check( x[i] );
    }
    for(int i = 0; i < 5; ++i )
    {
        if( i != 0 )
            putchar( ‘ ‘ );
        if( flag[i] )
        {
            if( i == 3 )
                printf( "%.1f", sum/res[i] );
            else
                printf( "%d", res[i] );
        }
        else
            putchar( ‘N‘ );
    }
    return 0;
}

代码君

1013. 数素数 (20)

打表然后输出，注意第i的含义，是第i个素数，所有表打大一点

#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

#define MAXN 2000000

int vis[ MAXN + 10 ], prime[ MAXN + 10 ];
int m, n, flag = 0, cnt, line = 0;

void init()
{
    int m = sqrt( MAXN + 0.5 );
    for( int i = 2; i <= m; ++i ) if( !vis[ i ] )
        for( int j = i * i; j <= MAXN; j += i )    vis[ j ] = 1;
    cnt = 0;
    for( int i = 2; i <= MAXN; ++i )
        if( !vis[i] )
            prime[ ++cnt ] = i;
}

int main()
{
    init();
    scanf( "%d %d", &n, &m );    
    for( int i = n; i <= m; ++i )
    {
        if( line == 10 )
        {
            putchar( ‘\n‘ );
            line = 1;
            flag = 0;
        }
        else
            line++;
        if( flag ) putchar( ‘ ‘ );
        else     flag = 1;
        printf( "%d", prime[i] );
    }    
    putchar( ‘\n‘ );
}

代码君

1014. 福尔摩斯的约会 (20)

注意先看清楚题目，然后按照题目意思来，注意输出格式

#include <cstdio>
#include <cstring>
using namespace std;

char str[4][62];
int DAY, HH, MM;
int i, j, ct = 0;
char day[7][4] = {"MON","TUE","WED","THU","FRI","SAT","SUN"};
int main()
{
    for( i = 0; i < 4; ++i )
        scanf( "%s", str[i] );

    for( i = 0; str[0][i] && str[1][i]; ++i )
    {
        if( (str[0][i] <= ‘G‘ && str[0][i] >= ‘A‘ ) && ( str[1][i] == str[0][i] ) )
        {
            DAY = str[1][i] - ‘A‘;
            break;
        }
    }
    for( i = i + 1; str[0][i] && str[1][i]; ++i )
    {
        if( ( str[0][i] <= ‘N‘ && str[0][i] >= ‘A‘ ) || ( str[0][i] <= ‘9‘ && str[0][i] >= ‘0‘ ) )
        {
            if( str[1][i] == str[0][i] )
            {
                HH = str[1][i];
                break;
            }
        }
    }
    for( i = 0; str[2][i] && str[3][i]; ++i )
    {
        if( ( str[2][i] <= ‘z‘ && str[2][i] >= ‘a‘ ) || ( str[2][i] <= ‘Z‘ && str[2][i] >= ‘A‘ ) )
        {
            if( str[2][i] == str[3][i] )
            {
                MM = i;
                break;
            }
        }
    }
    printf( "%s ",day[ DAY ] );
    if( HH > ‘9‘)   printf( "%d", HH - ‘A‘ + 10 );
    else            printf( "0%d", HH - ‘0‘ );
    if( MM >= 10 )   printf( ":%d\n", MM );
    else            printf( ":0%d\n", MM );
    return 0;
}

代码君

1015. 德才论 (25)

分个类，排个序

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define MAXN 100000

struct node
{
    int id;
    int ds;
    int cs;
};

bool cmp( node x, node y )
{
    if( y.ds + y.cs == x.ds + x.cs )
    {
        if( x.ds == y.ds )
            return x.id < y.id;
        return x.ds > y.ds; 
    }
    return x.ds + x.cs > y.ds + y.cs;
}

node A[MAXN], B[MAXN], C[MAXN], D[MAXN];
int N, L, H;
int x, y, z, ct = 0;
int ca, cb, cc, cd;

int main()
{
    scanf( "%d %d %d", &N, &L, &H );
    ca = cb = cc = cd = 0;
    for( int i = 0; i < N; ++i )
    {
        scanf( "%d %d %d", &x, &y, &z );
        if( y < L || z < L )
            continue;
        if( y >= H && z >= H )
        {
            A[ca].id = x;
            A[ca].ds = y;
            A[ca++].cs = z;
        }
        else if( y >= H && z < H )
        {
            B[cb].id = x;
            B[cb].ds = y;
            B[cb++].cs = z;
        }
        else if( y < H && z < H && y >= z )
        {
            C[cc].id = x;
            C[cc].ds = y;
            C[cc++].cs = z;
        }
        else
        {
            D[cd].id = x;
            D[cd].ds = y;
            D[cd++].cs = z;
        }
        ct++;
    }
    sort( A, A + ca, cmp );
    sort( B, B + cb, cmp );
    sort( C, C + cc, cmp );
    sort( D, D + cd, cmp );
    printf( "%d\n", ct );
    for( int i = 0 ; i < ca; ++i )
        printf( "%d %d %d\n", A[i].id, A[i].ds, A[i].cs );
    for( int i = 0 ; i < cb; ++i )
        printf( "%d %d %d\n", B[i].id, B[i].ds, B[i].cs );
    for( int i = 0 ; i < cc; ++i )
        printf( "%d %d %d\n", C[i].id, C[i].ds, C[i].cs );
    for( int i = 0 ; i < cd; ++i )
        printf( "%d %d %d\n", D[i].id, D[i].ds, D[i].cs );
    return 0;
}

代码君

1016. 部分A+B (15)

简单水题

#include <cstdio>
#include <cstring>
using namespace std;

char s1[20], s2[20];
int a, b, x, y;
int main()
{
    scanf( "%s %d %s %d", s1,  &a, s2, &b );
    x = y = 0;
    for( int i = 0 ; s1[i]; ++i )
    {
        if( s1[i] - ‘0‘ == a )
            x = x * 10 + a;
    }
    for( int i = 0 ; s2[i]; ++i )
    {
        if( s2[i] - ‘0‘ == b )
            y = y * 10 + b;
    }
    printf( "%d\n", x + y );
    return 0;
}

代码君

1017. A除以B (20)

模拟除法

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

char s[1010];
char r[1010];
int v, m, ct, i, len;

int main()
{
    scanf( "%s %d", s, &m );
    v = ct = 0;
    len = strlen( s );
    while( v < m )
    {
        v = v*10 + s[i] - ‘0‘;
        i++;
        if( i >= len )
        {
            printf( "0 %d", v );
            return 0;
        }
    }
    for( ; i < len; ++i )
    {
        r[ct++] = v/m + ‘0‘;
        v = v%m;
        v = v*10 + s[i] - ‘0‘;
    }
    r[ct++] = v/m + ‘0‘;
    v = v%m;
    printf( "%s %d", r, v );
    return 0;
}

代码君

1018. 锤子剪刀布 (20)

模拟题，注意标记

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

char A, B;
int n;
int W[2], E;

struct node
{
    int ct;
    char c;
    bool operator < ( const node b ) const
    {
        if( ct == b.ct )
            return c < b.c;
        return ct > b.ct; 
    }
}NA[3], NB[3];

int main()
{
    NA[0].c = NB[0].c = ‘C‘;
    NA[1].c = NB[1].c = ‘J‘;
    NA[2].c = NB[2].c = ‘B‘;
    scanf( "%d", &n );
    while( n-- )
    {
        getchar();
        scanf( "%c %c", &A, &B );
        if( A == B )    E++;
        else if( A == ‘C‘ && B == ‘J‘ ) W[0]++, NA[0].ct++;
        else if( A == ‘J‘ && B == ‘B‘ ) W[0]++, NA[1].ct++;
        else if( A == ‘B‘ && B == ‘C‘ ) W[0]++, NA[2].ct++;
        else if( B == ‘C‘ && A == ‘J‘ ) W[1]++, NB[0].ct++;
        else if( B == ‘J‘ && A == ‘B‘ ) W[1]++, NB[1].ct++;
        else if( B == ‘B‘ && A == ‘C‘ ) W[1]++, NB[2].ct++;
    }    
    
    sort( NA, NA + 3 );
    sort( NB, NB + 3 );
    printf( "%d %d %d\n", W[0], E, W[1] );
    printf( "%d %d %d\n", W[1], E, W[0] );
    printf( "%c %c\n", NA[0].c, NB[0].c );    
    return 0;
}

代码君

1019. 数字黑洞 (20)

简单模拟，注意字符串和数值的转换

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

char s1[5], s2[5], tp[5];
int x, y, z, num;

void tran( int x, char tar[])
{
    tar[0] = x/1000 + ‘0‘,    x %= 1000;
    tar[1] = x/100 + ‘0‘,    x %= 100;
    tar[2] = x/10 + ‘0‘;    x %= 10;
    tar[3] = x + ‘0‘;
}

void rev( char src[], char tar[] )
{
    int i, j;
    for( i = 3, j = 0; i >= 0; --i, ++j )
        tar[j] = src[i];
}
bool cmp( char x, char y )
{
    return x > y;
}

int main()
{
    scanf( "%d", &num );
    tran( num, s1 );
    if( s1[0] == s1[1] && s1[1] == s1[2] && s1[2] == s1[3] )
        printf( "%s - %s = 0000\n", s1, s1 );
    else
    {
        while( true )
        {
            sort( s1, s1 + 4, cmp );
            rev( s1, s2 );
            z = x = y = 0;
            for( int i = 0; i < 4; ++i )
            {
                x = x * 10 + s1[i] - ‘0‘;
                y = y * 10 + s2[i] - ‘0‘;                
            }
            z = x - y;
            tran( z, tp );
            printf( "%s - %s = %s\n", s1, s2, tp );
            if( z == 6174 )
            {
                break;
            }
            else 
            {
                tran( z, s1 );
                rev( s1, s2 );
            }                
        }
    }
    
    return 0;
}

代码君

1020. 月饼 (25)

简单贪心

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct node
{
    double tot;
    double val;
    double bit;
}E[1010];

bool cmp( node x, node y )
{
    return x.bit > y.bit;
}

int n, m;
double res;

int main()
{
    scanf( "%d %d", &n, &m );
    for( int i = 0; i < n; ++i )
        scanf( "%lf", &E[i].tot );
    for( int i = 0; i < n; ++i )
    {
        scanf( "%lf", &E[i].val );
        E[i].bit = E[i].val / E[i].tot;
    }
    sort( E, E + n, cmp );
    res = 0;
    for( int i = 0; i < n; ++i )
    {
        if( E[i].tot < m )
        {
            m -= E[i].tot;
            res += E[i].val;
        }
        else 
        {
            res += E[i].bit * m;
            break;
        }
    }
    printf( "%.2f\n", res );
    return 0;
}

代码君

1021. 个位数统计 (15)

水题

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int D[10];
char str[1010];

int main()
{
    scanf( "%s", str );
    for( int i = 0; str[i]; ++i )
        D[ str[i] - ‘0‘ ]++;
    for( int i = 0; i < 10; ++i )
    {
        if( D[i] != 0 )
            printf( "%d:%d\n", i, D[i] );
    }
    return 0;
}

代码君

1022. D进制的A+B (20)

进制转换，特判下0

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int A, B, C, D, ct = 0;
char res[ 100 ];

int main()
{
    scanf( "%d %d %d", &A, &B, &D );
    C = A + B;
    while( C )
    {
        res[ct++] = C % D + ‘0‘;
        C /= D;
    }
    
    if( ct == 0 )    
        putchar( ‘0‘ );
    else
    {
        for( int i = ct-1; i >=0 ; --i )
            printf( "%c", res[i] );
        putchar( ‘\n‘ ); 
    }
    return 0;
}

代码君

1023. 组个最小数 (20)

先找不是0的最小数，然后再由小到大排列

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int nc[10], t;

int main()
{
    for( int i = 0; i < 10 ; ++i )
        scanf( "%d", &nc[i] );
    t = 1;
    while( nc[t] == 0 ) t++;
    printf( "%d", t );
    nc[t]--;
    for( int i = 0; i < 10; ++i )
    {
        for( int j = 0; j < nc[i]; ++j )
            printf( "%d", i );
    }
    putchar( ‘\n‘ );
    return 0;
}

代码君

1024. 科学计数法 (20)

模拟，这题我代码写得有点撮

#include <cstdio>
#include <cstring>

char str[ 10010 ];
char res[ 10010 ];

int main()
{
    scanf( "%s", str );
    if( str[0] == ‘-‘ )    putchar( ‘-‘ );
    int t = strlen( str ) - 1;
    int flag = 0;
    while( str[t] != ‘E‘ ) t--;

    int ct = 0 ;
    for( int i = t + 2; str[i]; ++i )
    {
        ct = ct * 10 + str[i] - ‘0‘;
    }
    if( str[t+1] == ‘-‘ ) flag = 1;
    
    if( flag )
    {
        if( ct > 0 )
        {
            printf( "0." );
            for( int i = 1; i < ct; ++i )
                printf( "0" );
            for( int i = 1; i < t; ++i )
            {
                if( str[i] != ‘.‘ )
                    printf( "%c", str[i] );
            }
                
        }
        else
        {
            printf( "%c.", str[1] );
            for( int i = 2; i < t; ++i )
            {
                if( str[i] != ‘.‘ )
                    printf( "%c", str[i] );
            }
                
        }
    }
    else
    {
        int len = t - 3;
        if( ct >= len )
        {
            for( int i = 1; i < t; ++i )
            {
                if( str[i] != ‘.‘ )
                    printf( "%c", str[i] );    
            }
            for( int i = 0; i < ct - len; ++i )
                printf( "0" );
        }
        else if( ct == 0 )
        {
            for( int i = 1; i < t; ++i )
                printf( "%c", str[i] );    
        }        
        else
        {
            for( int i = 1; i < t; ++i )
            {
                if( str[i] != ‘.‘ )
                    printf( "%c", str[i] );    
                if( i == t - ct )
                    printf( "." );    
            }        
        }
    }
    putchar( ‘\n‘ );
     
    return 0;
}

代码君

1025. 反转链表 (25)

35道题目里面最难的3题之一，模拟链表

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct node
{
    int addr;
    int data;
    int next;
}D[ 100010 ], LIST[ 100010 ];

int start, n, k, add;

int main()
{
    scanf( "%d %d %d", &start, &n, &k );
    for( int i = 0; i < n; ++i )
    {
        scanf( "%d", &add );
        D[add].addr = add;
        scanf( "%d %d", &D[add].data, &D[add].next );
    }
    int ct = 0;
    while( start != -1 )
    {
        LIST[ct].addr = D[start].addr;
        LIST[ct].data = D[start].data;
        LIST[ct].next = D[start].next;
        start = D[start].next;
        ct++;
    }
    
    int be = 0, ed = k - 1;
    while( ed < ct )
    {
        while( ed > be )
        {
            LIST[ed].next = LIST[ed - 1].addr;
            printf( "%05d %d ", LIST[ed].addr, LIST[ed].data );  
            if( LIST[ed].next != -1 )
                printf( "%05d\n", LIST[ed].next );
            else
                printf( "-1\n" );
            ed--;
        }
        if( be + 2 * k - 1 < ct )
            LIST[be].next = LIST[be + 2 * k - 1].addr;
        else if( be + k == ct )
            LIST[be].next = -1;
        else
            LIST[be].next = LIST[be + k].addr;
        printf( "%05d %d ", LIST[be].addr, LIST[be].data );  
        
        if( LIST[be].next != -1 )  
            printf( "%05d\n", LIST[be].next );  
        else  
            printf( "%d\n", LIST[be].next );  
        be += k;        
        ed = be + k - 1;  
    }
    
    while( be < ct )
    {
        printf( "%05d %d ", LIST[be].addr, LIST[be].data );
        if( LIST[be].next != -1 )
            printf( "%05d\n", LIST[be].next );
        else
            printf( "-1\n" );
        be++;
    }
    
    return 0;
}

代码君

1026. 程序运行时间(15)

四舍五入一下，然后按格式输出即可

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int C1, C2, t;
int H, M, S;
int main()
{
    scanf( "%d %d", &C1, &C2 );
    t = (double)(C2 - C1)/100 + 0.5;
    H = t / 3600;
    t %= 3600;
    M = t/60;
    t %= 60;
    S = t;
    if( H < 10 )    printf( "0%d:", H );
    else            printf( "%d:", H );    
    if( M < 10 )    printf( "0%d:", M );
    else            printf( "%d:", M );    
    if( S < 10 )    printf( "0%d\n", S );
    else            printf( "%d\n", S );    
    return 0;
}

代码君

1027. 打印沙漏(20)

简单题，但是注意输出的格式，容易格式错误

#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

char ch;
int n, res, line;

int main()
{
    scanf( "%d %c", &n, &ch );
    line = sqrt( (n + 1) / 2 );
    res = n - line * line * 2 + 1;
    for( int i = line; i > 0; --i )
    {
        for( int k = line - i; k > 0; --k )
            putchar( ‘ ‘ );
        for( int j = 2*i-1; j > 0; --j )
            putchar( ch );
        putchar( ‘\n‘ );
    }
    for( int i = 2; i <= line ; ++i )
    {
        for( int k = line - i; k > 0; --k )
            putchar( ‘ ‘ );
        for( int j = 2*i-1; j > 0; --j )
            putchar( ch );
        putchar( ‘\n‘ );
    }
    printf( "%d\n", res );
    return 0;
}

代码君

1028. 人口普查(20)

此题略坑，，如果没有人，只输出0，注意特判

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct people
{
    char name[6];
    int year;
    int mon;
    int day;
}P[100010];

bool cmp( people x, people y )
{
    if( x.year == y.year )
    {
        if( x.mon == y.mon )
            return x.day < y.day;
        else
            return x. mon < y.mon;
    }
    else
        return x. year < y.year;
}

char s[6];
int y, m, d, n, ct = 0;

int main()
{
    scanf( "%d", &n );
    for( int i = 0; i < n; ++i )
    {
        scanf( "%s %d/%d/%d", s, &y, &m, &d );
        if( y >= 2015 )    continue;
        else if( y == 2014 && m > 9 ) continue;
        else if( y == 2014 && m == 9 && d > 6 ) continue;
        else if( y < 1814 )    continue;
        else if( y == 1814 && m < 9 ) continue;
        else if( y == 1814 && m == 9 && d < 6 ) continue;
        else
        {
            strcpy( P[ct].name, s );
            P[ct].year = y;
            P[ct].mon = m;
            P[ct++].day = d;         
        }
    }    
    sort( P, P + ct, cmp );
    if( ct )
        printf( "%d %s %s\n", ct, P[0].name, P[ct-1].name );
    else
        printf( "%d\n", ct );
        
    return 0;
}

代码君

1029. 旧键盘(20)

注意大小写的hash标记

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

char s1[100], s2[100];
int H[ 256 ], ct = 0;

void check()
{
    if( !H[ s1[ct] ] )
        {
            if( s1[ct] >= ‘A‘ && s1[ct] <= ‘Z‘ )
            {
                putchar( s1[ct] );
                H[ s1[ct] ] = 1;
                H[ s1[ct] - ‘A‘ + ‘a‘ ] = 1;
            }
            else if( s1[ct] >= ‘a‘ && s1[ct] <= ‘z‘ )
            {
                putchar( s1[ct] - ‘a‘ + ‘A‘ );
                H[ s1[ct] - ‘a‘ + ‘A‘ ] = 1;
                H[ s1[ct] ] = 1;
            }
            else
            {
                putchar( s1[ct] );
                H[ s1[ct] ] = 1;
            }
        }    
}
int main()
{
    scanf( "%s %s", s1, s2 );
    for( int i = 0; s2[i]; ++i )
    {
        for( ; s1[ct]; ++ct )
        {
            if( s1[ct] != s2[i] )
            {
                check();
            }
            else
            {
                ct++;
                break;
            }
                
        }
    }    
    while( s1[ct] )
    {
        check();
        ct++;
    }
    putchar( ‘\n‘ );
    return 0;
}

代码君

1030. 完美数列(25)

注意优化，小心超时，注意数据范围

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

long long x[ 100010 ], M, p;
int n, flag = 0, t, maxx = 0;

int main()
{
    scanf( "%d %lld", &n, &p );
    for( int i = 0; i < n; ++i )
        scanf( "%lld", &x[i] );    
    sort( x, x + n );
    for( int i = 0; i < n; ++i )
    {
        for( int j = i + maxx - 1; j < n; ++j )
        {
            if( j < 0 )    continue;
            if( x[i]*p < x[j] )    break;
            if( j - i + 1 > maxx )
                maxx = j - i + 1;
        }
    }
    printf( "%d\n", maxx );
    return 0;
}

代码君

1031. 查验身份证(15)

按题意模拟即可

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int n, sum = 0, ct = 0, flag = 0;
char s[20];
char M[15] = {"10X98765432"};
int V[20] = {7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2};

int main()
{
    scanf( "%d", &n );
    for( int i = 0; i < n; ++i )
    {
        scanf( "%s", s );
        flag = sum = 0;
        for( int i = 0; i < 17; ++i )
        {
            if( s[i] > ‘9‘ || s[i] < ‘0‘ )
            {
                flag = 1;
                break;
            }
            sum += ( s[i] - ‘0‘ )*V[i];
        }
        if( flag )
        {
            puts( s );
        }
        else
        {
            sum %= 11;
            if( M[sum] == s[17] )
                ct++;
            else
                puts( s );    
        }
    }
    if( ct == n )
        puts( "All passed" );
}

代码君

1032. 挖掘机技术哪家强(20)

简单水题

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int x[ 100010 ], n;
int H[ 100010 ], sch, sco, res = 0, tot = 0;

int main()
{
    scanf( "%d", &n );
    for( int i = 0; i < n; ++i )
    {
        scanf( "%d %d", &sch, &sco );
        x[ sch ] += sco;
        if( !H[ sch] )
        {
            tot++;
            H[ sch ] = 1; 
        }
    }
    
    for( int i = 1; i <= tot; ++i )
    {
        if( res < x[i] )
        {
            res = x[i];
            sch = i;
        }
    }
    printf( "%d %d\n", sch, x[sch] );
    
    return 0;
}

代码君

1033. 旧键盘打字(20)

这题需要用gets()，有空格，貌似题目没有说清楚啊。。。。这个是坑

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

char s1[ 256 ], s2[ 100010 ];
int H[ 256 ], shift = 0;


int main()
{
    gets( s1 );
    gets( s2 );
    for( int i = 0; s1[i]; ++i )
    {
        if( s1[i] >= ‘A‘ && s1[i] <= ‘Z‘ )        
        {
            H[ s1[i] ] = 1;
            H[ s1[i] - ‘A‘ + ‘a‘ ] = 1;
        }
        else if( s1[i] == ‘+‘ )
        {
            shift = 1;
        }
        else
        {
            H[ s1[i] ] = 1;
        }
    }    
    for( int i = 0; s2[i]; ++i )
    {
        if( s2[i] >= ‘A‘ && s2[i] <= ‘Z‘ && shift )    
            continue;
        if( H[ s2[i] ] )
            continue;
        putchar( s2[i] );
    }     
    putchar( ‘\n‘ );
    return 0;
}

代码君

1034. 有理数四则运算(20)

注意细节。先写好框架然后再调bug
题目保证正确的输出中没有超过整型范围的整数。但是不一定计算过程中没有超过，要小

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;

LL a1, a2, b1, b2, A, B;


LL ABS( LL x )
{
    if( x < 0 ) return  -x;
    return x;
}

LL gcd( LL a, LL b)
{
    return b == 0 ? a : gcd( b, a % b );
}

void out( LL x, LL y )
{
    LL GCD = gcd( x, y );
    x /= GCD;
    y /= GCD;
    if( x * y < 0 )    
    {
        printf( "(" );
        if( ( ( x / y >= 1 ) || ( x / y <= -1 ) ) && ( x % y == 0 ) )
            printf( "%lld", x / y );
        else if( x / y >= 1 || x / y <= -1 ) 
            printf( "%lld %lld/%lld", x / y, ABS( x % y ), ABS( y ) );
        else
            printf( "%lld/%lld", -ABS( x ) , ABS( y ) );    
        printf( ")" );
    }
    else
    {        
        if( x == 0 )
            printf( "0" );
        else if( ( ( x / y >= 1 ) || ( x / y <= -1 ) ) && ( x % y == 0 ) )
            printf( "%lld", x / y );
        else if( x / y >= 1 || x / y <= -1 ) 
            printf( "%lld %lld/%lld", x / y, x % y, y );
        else
            printf( "%lld/%lld", x , y );    
    }
}

void calc( LL a1, LL b1, LL a2, LL b2, char op )
{
    switch( op )
    {
        case ‘+‘:
            A = a1 * b2 + a2 * b1;
            B = b1 * b2;    
            out( a1, b1 );
            printf( " + " );
            out( a2, b2 );
            printf( " = " );
            out( A, B );
            putchar( ‘\n‘ );
            break;
        case ‘-‘:
            A = a1 * b2 - a2 * b1;
            B = b1 * b2;
            out( a1, b1 );
            printf( " - " );
            out( a2, b2 );
            printf( " = " );
            out( A, B );
            putchar( ‘\n‘ );
            break;
        case ‘*‘:
            A = a1 * a2;
            B = b1 * b2;    
            out( a1, b1 );
            printf( " * " );
            out( a2, b2 );
            printf( " = " );
            out( A, B );
            putchar( ‘\n‘ );
            break;
        case ‘/‘:
            A = a1 * b2;
            B = a2 * b1;    
            out( a1, b1 );
            printf( " / " );
            if( a2 == 0 )
            {
                printf( "0 = Inf\n");
            }
            else
            {
                out( a2, b2 );
                printf( " = " );
                out( A, B );
                putchar( ‘\n‘ );    
            }
            break;
    }
}


int main()
{
    scanf( "%lld/%lld %lld/%lld", &a1, &b1, &a2, &b2 );
    
    calc( a1, b1, a2, b2, ‘+‘ );
    calc( a1, b1, a2, b2, ‘-‘ );
    calc( a1, b1, a2, b2, ‘*‘ );
    calc( a1, b1, a2, b2, ‘/‘ );
    
    return 0;
}

代码君

1035. 插入与归并(25)

排序排一轮就比一次。

#include <cstdio>
#include <cstring>
using namespace std;


int x1 [ 110 ], x2[ 110 ], tar[ 110 ], buf[ 220 ], n;

void scan( int& x )
{
    char c;
    while( c = getchar(), c < ‘0‘ || c > ‘9‘ );
    x = c - ‘0‘;
    while( c = getchar(), c >= ‘0‘ && c <= ‘9‘ ) x = x * 10 + c - ‘0‘;
}

void Merge_Sort( int arr[], int n )
{
    int flag = 0, j;
    int step, lbe, led, rbe, red;

    for( step = 1; step < n; step <<= 1 )
    {
        for( lbe = 0; lbe < n - step; lbe = red + 1 )
        {
            rbe = lbe + step;
            led = rbe - 1;
            red = led + step;
            if( red > n - 1 )
                red = n - 1;
            int k = 0;
            memset( buf, 0, sizeof(buf) );
            
            while( lbe <= led && rbe <= red )
            {
                if( arr[lbe] < arr[rbe] )
                    buf[k++] = arr[lbe++];
                else
                    buf[k++] = arr[rbe++];
            }
            while( lbe <= led )    buf[k++] = arr[lbe++];
            while( rbe <= red )    buf[k++] = arr[rbe++];
            while( k > 0 )        arr[ --rbe ] = buf[ --k ];
        }
        if( flag )
        {
            printf( "Merge Sort\n" );
            for( j = 0; j < n; ++j )
            {
                if( j != 0 ) putchar( ‘ ‘ );
                printf( "%d", arr[j] );
            }
            putchar( ‘\n‘ );
            return;
        }
        for( j = 0; j < n; ++j )
        {
            if( arr[j] != tar[j] )
                break;
        }
        if( j == n )
            flag = 1;
    }
}

bool Insert_Sort( int arr[], int n )
{
    int b, e, t, i, flag = 0;
    
    for( b = 1; b < n; ++b )
    {
        t = arr[b];
        e = b;
        while( e > 0 && arr[ e - 1 ] > t )
        {
            arr[ e ] = arr[ e - 1 ];
            --e;
        }
        arr[ e ] = t;    
        
        for( i = 0; i < n; ++i )
        {
            if( arr[i] != tar[i] )
                break;
        }
        if( flag )
        {
            printf( "Insertion Sort\n" );
            for( i = 0; i < n; ++ i )
            {
                if( i != 0 )    putchar( ‘ ‘ );
                printf( "%d", arr[i] );
            }
            putchar( ‘\n‘ );
            return true;
        }
        if( i == n )
            flag = 1;
    }
    return false;
}


int main()
{
    scan( n );
    for( int i = 0; i < n; ++i )
    {
        scan( x1[i] );
        x2[i] = x1[i];
    }
    for( int i = 0; i < n; ++i )
        scan( tar[i] );
    if( !Insert_Sort( x1, n ) )
        Merge_Sort( x2, n );
        
    return 0;
}

代码君

PAT从零单排

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原文地址：http://www.cnblogs.com/BigBallon/p/4229907.html

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