UESTC 2014 Summer Training #19

时间：2014-08-10 01:32:59 阅读：340 评论：0 收藏：0 [点我收藏+]

　　去迟了,队友已经开始写了,应该是个水题,贴个队友代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<stack>
#include<queue>
using namespace std;
#define INF 1000000000
#define eps 1e-8
#define pii pair<int,int>
#define LL long long int
int n;
char s[1000000];
int main()
{
    //freopen("in2.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    scanf("%d",&n);
    while(n--)
    {
        //memset(s,0,sizeof(s));
        scanf("%s",s);
        int t=strlen(s);
        if(s[t/2]==s[t/2-1])
        {
            cout<<"Do-it"<<endl;
        }
        else
        {
            cout<<"Do-it-Not"<<endl;
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}

B.UVALive 6162

　　简单模拟:检查每个记录是否超过2h,记录时间的时候注意是有大部分时间在夜晚则全部记录为夜行时间(分类讨论下)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

int n, a, b, sunrise, sunset, st, et, cost, nt, s1, s2;

int toMinut(int h, int s)
{
    return h*60+s;
}

int main()
{
#ifdef LOCAL
    freopen("B.in", "r", stdin);
#endif
    while(scanf("%d", &n)) {
        if(n == 0)    break;
        s1 = s2 = 0;
        bool ok = true;
        for(int i = 0; i < n; i++) {
            scanf("%d:%d", &a, &b);
            sunrise = toMinut(a, b);
            scanf("%d:%d", &a, &b);
            sunset = toMinut(a, b);
            scanf("%d:%d", &a, &b);
            st = toMinut(a, b);
            scanf("%d:%d", &a, &b);
            et = toMinut(a, b);
            //finish input
            //check if more than 2 hours
            cost = et - st;
            if(cost > 120)    ok = false;
            //check if drive in night
            if(st < sunrise && et <= sunset) {
                nt = sunrise - st;
                if(nt*2 >= cost)    s2 += cost;
            }
            else if(et > sunset && st >= sunset) {
                nt = et - sunset;
                if(nt*2 >= cost)    s2 += cost;
            }
            else if(st >= sunrise && et <= sunset)
                s2 += cost;
            else if(st <= sunrise && et >= sunset) {
                nt = et-sunset + sunrise- st;
                if(nt*2 >= cost)    s2 += cost; 
            }
            else    s1 += cost;
        }
        if(s1+s2 >= 3000 && s2 >= 600 && ok)    printf("PASS\n");
        else    printf("NON\n");
    }
    return 0;
}

C.UVALive 6163

　　暴力枚举每一种情况,可以想到一种分类方式:两个数‘运算‘,其结果与第三个数’运算‘,其结果与第四个数‘运算’;(两个数‘运算‘)‘运算’(两个数‘运算‘) (后者我一开始漏掉了)
　　这样的话,枚举每一种排列,枚举上述方式,枚举每一个符号,枚举第一类时,A?B, B?A都要算进去

　　第一类用dfs枚举,第二类用三重循环,见代码

　　感觉还是比较容易漏情况的...

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<stack>
#include<queue>
using namespace std;
#define INF 1000000000
#define eps 1e-8
#define pii pair<int,int>
#define LL long long int
int n,a[4];
char s[4]={‘+‘,‘-‘,‘*‘,‘/‘};
char in[4];
bool ok;
bool deal();
void dfs();
int main()
{
    //freopen("out.txt","w",stdout);
    while(scanf("%d",&n)==1&&n)
    {
        bool ans=true;
        for(int i=1;i<=n;i++)
        {
            scanf("%s",in);
            if(ans==false)
                continue;
            a[0]=in[0]-‘0‘;
            a[1]=in[1]-‘0‘;
            a[2]=in[2]-‘0‘;
            a[3]=in[3]-‘0‘;
            if(deal()==false)//这四个数弄不出10
            {
                ans=false;
            }
        }
        if(ans==true)
            printf("TRUE\n");
        else
            printf("BUSTED\n");
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}
void dfs(int x, int ans)
{
//    cout << x << " " << ans << endl;
    if(x == 4 && ans == 10) {
        ok = true;
        return;
    }
    if(ok || x >= 4)  return;
    int c = ans, d = a[x];
    dfs(x+1, c+d);
    dfs(x+1, c-d);
    dfs(x+1, d-c);
    dfs(x+1, c*d);
    if(d != 0)
        dfs(x+1, c/d);
    if(c != 0)
    dfs(x+1, d/c);
}
int calc(int x, int y, int t)
{
    switch(t) {
        case 0: return x+y;
        case 1: return x-y;
        case 2: return x*y;
        case 3: return x/y;
        default:    return -1;
    }
}
bool deal()//看目前的四个数能不能弄出10
{
    ok = false;
sort(a, a+4);
    do
    {
        dfs(1, a[0]);
        for(int i = 0; i < 4; i++) {
            if(a[1] == 0 && i == 3) continue;
            int x = calc(a[0], a[1], i);
            for(int j = 0; j < 4; j++) {
                if(a[3] == 0 && j == 3) continue;
                int y = calc(a[2], a[3], j);
                for(int k = 0; k < 4; k++) {
                    if(y == 0 && k == 3)    continue;
                    if(calc(x, y, k) == 10) {
                        ok = true;
                        break;
                    }
                }
            }
        }
        if(ok)  return true;
    }while(next_permutation(a,a+4));
    return false;
}