Trouble

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

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

1 -1

1 2 3

-1 -2 -3

4 5 6

-1 3 2

-4 -10 -1

Sample Output

Yes

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 ;

}

巴特西

HDU 4334 Trouble (数组合并)

Trouble

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