hdu 4334 Trouble

Trouble

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4375    Accepted Submission(s): 1297

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

 

 

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <set>
using namespace std;
#define ll long long
ll a[5][300],s1[90000],s2[90000],s3[90000],sn1,sn2,sn3;
int main()
{
    int n,t,i,j,k;
    cin>>t;
    while(t--)
    {
        cin>>n;
        for(i=0; i<5; i++)
            for(j=0; j<n; j++)
                scanf("%I64d",&a[i][j]);
        sn1=sn2=sn3=0;
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
                s1[sn1++]=a[0][i]+a[1][j];
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
                s2[sn2++]=a[2][i]+a[3][j];
        for(i=0; i<n; i++)
            s3[sn3++]=-a[4][i];
        sort(s1,s1+sn1);
        sort(s2,s2+sn2);
        sort(s3,s3+sn3);
        int ok=0;
        for(i=0;!ok&&i<sn3;i++)
        {
            j=0,k=sn2-1;
            while(j<sn1&&k>=0)
            {
                if(s1[j]+s2[k]==s3[i])
                {
                    ok=1;
                    break;
                }
                while(j<sn1&&k>=0&&s1[j]+s2[k]>s3[i])k--;
                while(j<sn1&&k>=0&&s1[j]+s2[k]<s3[i])j++;
            }
        }
        if(ok)printf("Yes
");
        else printf("No
");
    }
}