HDU 4334

Trouble

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2961 Accepted Submission(s): 893

Problem Description

Hassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him.
The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0?

 

Input

First line of input contains a single integer N (1≤N≤50). N test-cases follow. First line of each test-case contains a single integer n (1<=n<=200). 5 lines follow each containing n integer numbers in range [-10^15, 1 0^15]. I-th line denotes set S_i for 1<=i<=5.

 

Output

For each test-case output "Yes" (without quotes) if there are a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0, otherwise output "No".

 

Sample Input

2 2 1 -1 1 -1 1 -1 1 -1 1 -1 3 1 2 3 -1 -2 -3 4 5 6 -1 3 2 -4 -10 -1

 

Sample Output

No Yes

//大致题意：判断五个数之和是否可能为0 
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

typedef __int64 LL;//用define时出了错误 
LL a[210];
LL b[210];
LL c[210];
LL f[202*202];
LL g[202*202];
int num;

int main()
{
    int i,j,k,T;
    cin>>T;
    while(T --)
    {
        cin>>num;
        //memset(f, 0, sizeof f);
        //memset(g, 0, sizeof g);
        int cnt1 = 0;
        int cnt2 = 0;
        for(i = 0; i < num; i ++) 
          cin>>a[i];
        for(i = 0; i < num; i ++) 
          cin>>b[i];
        for(i = 0; i < num; i ++)
            for(j = 0; j < num; j ++)
                f[cnt1++] = a[i] + b[j];
        for(i = 0; i < num; i ++) 
          cin>>a[i];
        for(i = 0; i < num; i ++) 
          cin>>b[i];
        for(i = 0; i < num; i ++)
            for(j = 0; j < num; j ++)
                g[cnt2++] = a[i] + b[j];
        for(i = 0; i < num; i ++) 
          cin>>c[i];
        sort(f, f + cnt1);
        sort(g, g + cnt2);
        sort(c, c + num);
        bool flag = 0;
        for(i = 0; i < num; i ++)
        {
            LL temp = c[i];
            j = 0;
            k = cnt2 - 1;
            while(j < cnt1 && k >= 0)
            {
                if(f[j] + g[k] == (-temp))
                {
                    flag = 1;
                    break;
                }
                if(f[j] + g[k] <(-temp)) 
                    j++;
                else 
                    k--;
            }
        }
        if(flag)
          cout<<"Yes"<<endl;
        else
          cout<<"No"<<endl;
    }
    return 0;
}