2016"百度之星" - 资格赛 解题报告

/*
Problem :
Status  :

By wf,
*/

#include "algorithm"
#include "iostream"
#include "cstring"
#include "cstdio"
#include "string"
#include "stack"
#include "cmath"
#include "queue"
#include "set"
#include "map"

#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1

typedef long long ll;
using namespace std;

const int inf=0x3f3f3f3f;
const int mod=9973;

const int maxn = 1e6+5;

char str[1000000+5];
int sum[maxn<<2];
int index;
int len;

void PushUP(int rt)
{
    sum[rt] =( sum[rt<<1] * sum[rt<<1|1] )%mod;
}
void build(int l,int r,int rt)
{
    if(l==r)
    {

        //printf("%c ",str[index]);
        sum[rt] = str[index++]-28;
    }

    if (l == r) return ;
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUP(rt);
}

int query(int L,int R,int l,int r,int rt)
{
    if (L <= l && r <= R)
    {
        return sum[rt];
    }
    int m = (l + r) >> 1;
    int ret = 1;
    if (L <= m) ret = (ret * query(L , R , lson) )%mod ;
    if (R > m) ret = (ret* query(L , R , rson) ) %mod;
    //printf("ret==%d\n",ret);
    return ret;
}

int main()
{
    //freopen("in_a.txt","r",stdin);
    int n,a,b;
    int ans = 0;
    while (~scanf("%d",&n))
    {
        index=0;
        scanf("%s",&str);
        len = strlen(str);
        build(1 , len  , 1);

        int a,b;
        for(int i=0;i<n;++i)
        {
            scanf("%d %d",&a,&b);
            if(a>=1 && a<=len && 1<=b && b<=len) ans = query(a,b,1,len,1);
            printf("%d\n", query(a,b,1,len,1) );
        }


    }

    return 0;
}

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
/*  * 完全大数模板  * 输出cin>>a  * 输出a.print();  * 注意这个输入不能自动去掉前导0的，可以先读入到char数组，去掉前导0，再用构造函数。  */
#define MAXN 9999
#define MAXSIZE 1010
#define DLEN 4
class BigNum
{
private:
    int a[500];  //可以控制大数的位数
    int len;
public:
    BigNum()
    {
        len=1;    //构造函数
        memset(a,0,sizeof(a));
    }
    BigNum(const int);     //将一个int类型的变量转化成大数
    BigNum(const char*);   //将一个字符串类型的变量转化为大数
    BigNum(const BigNum &); //拷贝构造函数
    BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算
    friend istream& operator>>(istream&,BigNum&); //重载输入运算符
    friend ostream& operator<<(ostream&,BigNum&); //重载输出运算符
    BigNum operator+(const BigNum &)const;  //重载加法运算符，两个大数之间的相加运算
    BigNum operator-(const BigNum &)const;  //重载减法运算符，两个大数之间的相减运算
    BigNum operator*(const BigNum &)const;  //重载乘法运算符，两个大数之间的相乘运算
    BigNum operator/(const int &)const;     //重载除法运算符，大数对一个整数进行相除 运算
    BigNum operator^(const int &)const;     //大数的n次方运算

    int operator%(const int &)const;        //大数对一个int类型的变量进行取模运算
    bool operator>(const BigNum &T)const;   //大数和另一个大数的大小比较
    bool operator>(const int &t)const;      //大数和一个int类型的变量的大小比较
    void print();        //输出大数
};
BigNum::BigNum(const int b)   //将一个int类型的变量转化为大数
{
    int c,d=b;
    len=0;
    memset(a,0,sizeof(a));
    while(d>MAXN)
    {
        c=d-(d/(MAXN+1))*(MAXN+1);
        d=d/(MAXN+1);
        a[len++]=c;
    }
    a[len++]=d;
}
BigNum::BigNum(const char *s)  //将一个字符串类型的变量转化为大数
{
    int t,k,index,L,i;
    memset(a,0,sizeof(a));
    L=strlen(s);
    len=L/DLEN;
    if(L%DLEN)len++;
    index=0;
    for(i=L-1; i>=0; i-=DLEN)
    {
        t=0;
        k=i-DLEN+1;
        if(k<0)k=0;
        for(int j=k; j<=i; j++)             t=t*10+s[j]-‘0‘;
        a[index++]=t;
    }
}
BigNum::BigNum(const BigNum &T):len(T.len)  //拷贝构造函数
{
    int i;
    memset(a,0,sizeof(a));
    for(i=0; i<len; i++)         a[i]=T.a[i];
}
BigNum & BigNum::operator=(const BigNum &n)  //重载赋值运算符，大数之间赋值运算
{
    int i;
    len=n.len;
    memset(a,0,sizeof(a));
    for(i=0; i<len; i++)         a[i]=n.a[i];
    return *this;
}
istream& operator>>(istream &in,BigNum &b)
{
    char ch[MAXSIZE*4];
    int i=-1;
    in>>ch;

    int L=strlen(ch);
    int count=0,sum=0;
    for(i=L-1; i>=0;)
    {
        sum=0;
        int t=1;
        for(int j=0; j<4&&i>=0; j++,i--,t*=10)
        {
            sum+=(ch[i]-‘0‘)*t;
        }
        b.a[count]=sum;
        count++;
    }
    b.len=count++;
    return in;
}
ostream& operator<<(ostream& out,BigNum& b)  //重载输出运算符
{
    int i;
    cout<<b.a[b.len-1];
    for(i=b.len-2; i>=0; i--)
    {
        printf("%04d",b.a[i]);
    }
    return out;
}
BigNum BigNum::operator+(const BigNum &T)const   //两个大数之间的相加运算
{
    BigNum t(*this);
    int i,big;
    big=T.len>len?T.len:len;
    for(i=0; i<big; i++)
    {
        t.a[i]+=T.a[i];
        if(t.a[i]>MAXN)
        {
            t.a[i+1]++;
            t.a[i]-=MAXN+1;
        }
    }
    if(t.a[big]!=0)        t.len=big+1;
    else t.len=big;
    return t;
}
BigNum BigNum::operator-(const BigNum &T)const  //两个大数之间的相减运算
{
    int i,j,big;
    bool flag;
    BigNum t1,t2;
    if(*this>T)
    {
        t1=*this;
        t2=T;
        flag=0;
    }
    else
    {

        t1=T;
        t2=*this;
        flag=1;
    }
    big=t1.len;
    for(i=0; i<big; i++)
    {
        if(t1.a[i]<t2.a[i])
        {
            j=i+1;
            while(t1.a[j]==0)                 j++;
            t1.a[j--]--;
            while(j>i)                 t1.a[j--]+=MAXN;
            t1.a[i]+=MAXN+1-t2.a[i];
        }
        else t1.a[i]-=t2.a[i];
    }
    t1.len=big;
    while(t1.a[len-1]==0 && t1.len>1)
    {
        t1.len--;
        big--;
    }
    if(flag)         t1.a[big-1]=0-t1.a[big-1];
    return t1;
}
BigNum BigNum::operator*(const BigNum &T)const  //两个大数之间的相乘
{
    BigNum ret;
    int i,j,up;
    int temp,temp1;
    for(i=0; i<len; i++)
    {
        up=0;
        for(j=0; j<T.len; j++)
        {
            temp=a[i]*T.a[j]+ret.a[i+j]+up;
            if(temp>MAXN)
            {
                temp1=temp-temp/(MAXN+1)*(MAXN+1);
                up=temp/(MAXN+1);
                ret.a[i+j]=temp1;
            }
            else
            {
                up=0;
                ret.a[i+j]=temp;
            }
        }
        if(up!=0)            ret.a[i+j]=up;
    }
    ret.len=i+j;
    while(ret.a[ret.len-1]==0 && ret.len>1)ret.len--;
    return ret;

}
BigNum BigNum::operator/(const int &b)const  //大数对一个整数进行相除运算
{
    BigNum ret;
    int i,down=0;
    for(i=len-1; i>=0; i--)
    {
        ret.a[i]=(a[i]+down*(MAXN+1))/b;
        down=a[i]+down*(MAXN+1)-ret.a[i]*b;
    }
    ret.len=len;
    while(ret.a[ret.len-1]==0 && ret.len>1)         ret.len--;
    return ret;
}
int BigNum::operator%(const int &b)const   //大数对一个 int类型的变量进行取模
{
    int i,d=0;
    for(i=len-1; i>=0; i--)         d=((d*(MAXN+1))%b+a[i])%b;
    return d;
}
BigNum BigNum::operator^(const int &n)const  //大数的n次方运算
{
    BigNum t,ret(1);
    int i;
    if(n<0)exit(-1);
    if(n==0)return 1;
    if(n==1)return *this;
    int m=n;
    while(m>1)
    {
        t=*this;
        for(i=1; (i<<1)<=m; i<<=1)            t=t*t;
        m-=i;
        ret=ret*t;
        if(m==1)ret=ret*(*this);
    }
    return ret;
}
bool BigNum::operator>(const BigNum &T)const    //大数和另一个大数的大小比较
{
    int ln;
    if(len>T.len)return true;
    else if(len==T.len)
    {
        ln=len-1;
        while(a[ln]==T.a[ln]&&ln>=0)           ln--;
        if(ln>=0 && a[ln]>T.a[ln])            return true;
        else            return false;
    }
    else        return false;
}

bool BigNum::operator>(const int &t)const  //大数和一个int类型的变量的大小比较
{
    BigNum b(t);
    return *this>b;
}
void BigNum::print()   //输出大数
{
    int i;
    printf("%d",a[len-1]);
    for(i=len-2; i>=0; i--)       printf("%04d",a[i]);
    printf("\n");
}
BigNum f[250];//卡特兰数
int main()
{
    //freopen("in_b.txt","r",stdin);
    f[1]="1";
    f[2]="2";
    for(int i=3; i<=210; i++) f[i]=f[i-1]+f[i-2];
    int n;
    while(cin>>n)
    {
        if(n==0)
        {
            cout<<endl;
            continue;
        }
        cout<<f[n]<<endl;
    }
    return 0;
}

/*
Problem :
Status  :

By wf,
*/

#include "algorithm"
#include "iostream"
#include "cstring"
#include "cstdio"
#include "string"
#include "stack"
#include "cmath"
#include "queue"
#include "set"
#include "map"

#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1

typedef long long ll;
using namespace std;

const int inf=0x3f3f3f3f;
const int maxn=1e5+5;

#define MAX 26
struct Trie
{
    Trie *next[MAX];
    int v;   //根据需要变化

    Trie()
    {
        v=0;
        for(int i=0; i<MAX; ++i)
        {
            next[i]=NULL;
        }
    }
};

Trie *root;

void Insert(char *str)
{
    //printf("insert:%s\n",str);
    int len = strlen(str);
    Trie *p = root, *q;
    for(int i=0; i<len; ++i)
    {
        int id = str[i]-‘a‘;
        if(p->next[id]==NULL)
            p->next[id]=new Trie();
        p=p->next[id];
        p->v++;
    }
}
int Search(char *str)
{
    //printf("search:%s\n",str);
    int len = strlen(str);
    Trie *p = root;
    for(int i=0; i<len; ++i)
    {
        int id = str[i]-‘a‘;
        p = p->next[id];
        if(p == NULL)   //若为空集，表示不存以此为前缀的串
            return 0;

        //printf("%c v:%d\n",str[i],p->v);
    }
    return p->v;
}
void Delete(char * str)
{
    //printf("delete:%s\n",str);
    int len = strlen(str);

    //取删除单词的个数
    int num = Search(str);
    Trie *p = root;
    /*for(int i=0; i<len; ++i)
    {
        int id = str[i]-‘a‘;
        p = p->next[id];
        if(p == NULL)   //若为空集，表示不存以此为前缀的串
            break ;
        //printf("->%c ",str[i]);
        num = min(num,p->v) ;
    }*/

    //printf("num==%d\n",num);

    //这次再删
    p = root;
    for(int i=0; i<len; ++i)
    {
        int id = str[i]-‘a‘;
        p = p->next[id];
        if(p == NULL)   //若为空集，表示不存以此为前缀的串
            return ;
        //printf("->%c ",str[i]);
        p->v -= num;
        if(p->v ==0)
        {
            for(int i=0; i<26; ++i)
            {
                p->next[i]=NULL;
            }
            p=NULL;
            return;
        }
    }
    return ;
}



int main()
{
    //freopen("in_c.txt","r",stdin);
    int n;
    scanf("%d",&n);
    char cmd[20],str[5000];
    int len;
    root = new Trie();
    while(n--)
    {
        //cin>>cmd>>str;
        scanf("%s %s",&cmd,&str);
        switch( cmd[0] )
        {
        case ‘i‘:
            Insert(str);
            break;
        case ‘d‘:
            Delete(str);
            break;
        case ‘s‘:
            if( Search(str) )
            {
                cout<<"Yes"<<endl;
            }
            else cout<<"No"<<endl;
            break;
        }
    }

    return 0;
}

/*
Problem :
Status  :

By wf,
*/

#include "algorithm"
#include "iostream"
#include "cstring"
#include "cstdio"
#include "string"
#include "stack"
#include "cmath"
#include "queue"
#include "set"
#include "map"

#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1

typedef long long ll;
using namespace std;

const int inf=0x3f3f3f3f;
const int maxn=1e5+5;

map<string,int>mp;
int main()
{
    //freopen("in_d.txt","r",stdin);
    int n;
    string str;
    cin>>n;
    while(n--)
    {
        cin>>str;
        sort(str.begin(),str.end() );
        cout<<mp[str]<<endl;

            mp[str]++;

    }

    return 0;
}

/*
Problem :
Status  :

By wf,
*/

#include "algorithm"
#include "iostream"
#include "cstring"
#include "cstdio"
#include "string"
#include "stack"
#include "cmath"
#include "queue"
#include "set"
#include "map"

#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1

typedef long long ll;
using namespace std;

const int inf=0x3f3f3f3f;
const int maxn=1e5+5;

struct node
{
    string name;
    int l,r;
} var[50005];

struct node2
{
    int num;
    int varid[50];
    bool flag;
} p[1005];

map<string,int>mp;

bool iskong(int l1,int r1,int l2,int r2)
{
    if( l2>r1 || l1>r2 )return 1;
    return 0;
}

int main()
{
    //freopen("in_e.txt","r",stdin);
    string str;
    int n;
    cin>>n;
    getchar();
    int id = 1;
    for(int ii=0; ii<n; ii++)
    {
        getline(cin,str);
        //cout<<str<<endl;
        int i=0,j,len = str.length();
        p[ii].num=0;
        p[ii].flag=false;

        while(i<len)
        {
            while(i<len && str[i]==‘ ‘ )i++;
            if( isalpha(str[i]) )
            {
                //获取变量名
                j=i+1;
                while(j<len && isalpha(str[j]) )++j;
                var[id].name = str.substr(i,j-i);
                i=j;

                //获取运算符
                int ops=0;
                while(i<len && str[i]==‘ ‘ )i++;
                if( str[i]==‘<‘ && str[i+1]!=‘=‘ )
                {
                    ops=1;
                    i++;
                }
                else if( str[i]==‘>‘ && str[i+1]!=‘=‘ )
                {
                    ops=2;
                    i++;
                }
                else if( str[i]==‘<‘ && str[i+1]==‘=‘ )
                {
                    ops=3;
                    i+=2;
                }
                else if( str[i]==‘>‘ && str[i+1]==‘=‘ )
                {
                    ops=4;
                    i+=2;
                }
                else if( str[i]==‘=‘ && str[i+1]==‘=‘ )
                {
                    ops=5;
                    i+=2;
                }

                //获取数字
                int num=0;
                while(i<len && str[i]==‘ ‘)i++;
                if( isdigit(str[i]) )
                {
                    num = str[i]-‘0‘;
                    j=i+1;
                    while(j<len && isdigit(str[j]) )
                    {
                        num = num*10 + str[j] - ‘0‘;
                        j++;
                    }
                    i=j;
                }

                //处理
                switch(ops)
                {
                case 1: //<
                    var[id].l = -10000;
                    var[id].r = num-1;
                    break;
                case 2: //>
                    var[id].l = num+1;
                    var[id].r = 10000;
                    break;
                case 3: //<=
                    var[id].l = -10000;
                    var[id].r = num;
                    break;
                case 4: //>=
                    var[id].l = num;
                    var[id].r = 10000;
                    break;
                case 5: //==
                    var[id].l = num;
                    var[id].r = num;
                    break;
                }

                //判断这个条件是否为空
                //printf("dangqian num:%d\n",p[ii].num);
                for(int t=0; t<p[ii].num; t++)
                {
                    int tid = p[ii].varid[t];
                    //cout<<"name1:"<<var[id].name<<" name2:"<<var[tid].name<<endl;
                    if( var[id].name == var[tid].name && iskong(var[tid].l,var[tid].r,var[id].l,var[id].r) )
                    {
                        p[ii].flag=1;
                        //printf("debug\n");
                    }else
                    {
                        var[id].l = max( var[id].l , var[tid].l );
                        var[tid].l = var[id].l;
                        var[id].r = min( var[id].r , var[tid].r );
                        var[tid].r = var[id].r;
                    }
                }


                //处理完毕
                //cout<<name<<" ops=="<<ops<<" num=="<<num<<" l=="<<var[id].l<<" r=="<<var[id].r<<" flag=="<<p[ii].flag<<endl;

                p[ii].varid[ p[ii].num++]=id;
                id++;

                while(i<len && str[i]==‘ ‘)i++;
                if( i<len && str[i]==‘,‘ )
                {
                    i++;
                }

            }
            else i++;
        }

        if( p[ii].flag )
        {
            cout<<"unique"<<endl;
            continue;
        }

        bool isUnique=true;
        bool first = 1;

        for(int jj=0; jj<ii; ++jj)
        {
            bool isThisOk =1;
            if( p[jj].flag )continue;
            else
            {
                for(int k1 = 0; k1<p[ii].num; k1++)
                {
                    for(int k2 = 0; k2<p[jj].num; k2++)
                    {
                        if(!isThisOk )break;
                        int tid1 = p[ii].varid[k1];
                        int tid2 = p[jj].varid[k2];

                        //printf("[%d,%d] [%d,%d]\n",var[tid1].l,var[tid1].r,var[tid2].l,var[tid2].r);
                        //cout<<"name1:"<<var[tid1].name<<" name2:"<<var[tid2].name<<endl;
                        if( var[tid1].name == var[tid2].name && iskong(var[tid1].l,var[tid1].r,var[tid2].l,var[tid2].r) )
                        {
                            isThisOk=0;
                            //printf("ok\n");
                            break;
                        }
                    }
                }
            }

            if( isThisOk )
            {
                isUnique=false;
                if(first)first=0;
                else cout<<" ";
                cout<<jj+1;
            }

        }
        if( isUnique )cout<<"unique"<<endl;
        else cout<<endl;

        //cout<<endl;
    }
    return 0;
    return 0;
}