Trouble

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5681    Accepted Submission(s):
1570

Problem DescriptionHassan 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? InputFirst 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. OutputFor 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 Input221 -11 -11 -11 -11 -131 2 3-1 -2 -34 5 6-1 3 2-4 -10 -1Sample OutputNoYes题意：t组数据，每组输入一个n，接下来有五行代表五个集合，每行n个数，问是否能从每个集合中找出一个数，　　　使得和为0。题解：直接暴力会超时，将12两个集合合并，34两个集合合并，然后对他们进行从小到大排序，跑第五个集合。　　　用两个指针下标将12从小到大遍历，34逆序从大到小遍历，如果和<0，12集合的指针后移，如果和>0，　　　34集合的指针左移，如果和为0标记flag=1;break一下输出就行了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
long long a[],b[],c[],d[],e[],f[],g[];
long long i,j,x,k,l,n,t,flag=,p;
int main()
{
    scanf("%lld",&t);
    while(t--)
    {
        flag=;
        //输入
        scanf("%lld",&n);
        for(i=;i<n;i++)
        scanf("%lld",&d[i]);
        for(i=;i<n;i++)
        scanf("%lld",&e[i]);
        for(i=;i<n;i++)
        scanf("%lld",&f[i]);
        for(i=;i<n;i++)
        scanf("%lld",&g[i]);
        for(i=;i<n;i++)
        scanf("%lld",&c[i]);
        //合并
        int cnt1=;
        for(i=;i<n;i++)
            for(j=;j<n;j++)
            a[cnt1++]=d[i]+e[j];
        int cnt2=;
        for(i=;i<n;i++)
            for(j=;j<n;j++)
            b[cnt2++]=f[i]+g[j];  //三四合并
        sort(a,a+cnt1);
        sort(b,b+cnt2);
        //计算
        for(i=;i<n;i++) //第五个数组 c[]
        {
            j=;
            k=cnt2-;
             while(j<cnt1 && k>=)
             {
                 if(a[j]+b[k]+c[i]==0LL)
                 {
                     flag=;
                     break;
                 }
                 else if(a[j]+b[k]+c[i]<)  //注意不要都写成i
                     j++;
                 else
                     k--;
             }        }
        if(flag) cout<<"Yes"<<endl; //注意是Yes不是YES也不是yes
        else cout<<"No"<<endl;
    }
    return ;
}