poj 2785

4 Values whose Sum is 0

Time Limit: 15000MS Memory Limit: 228000KTotal Submissions: 18686 Accepted: 5546Case Time Limit: 5000MS

Description

The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 2²⁸ ) that belong respectively to A, B, C and D .

Output

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input

6-45 22 42 -16-41 -27 56 30-36 53 -37 77-36 30 -75 -4626 -38 -10 62-32 -54 -6 45

Sample Output

5

Hint

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

Source

Southwestern Europe 2005

注意双向枚举，否则单向处理的话会超时。

#include<cstdio>

#include<cstring>

#include<algorithm>

  

using namespace std;

  

int n;

const int N=5000;

int a[N],b[N],c[N],d[N];

int cd[N*N];

int main()

{

    while(scanf("%d",&n)!=EOF)

    {

        for(inti=0;i<n;i++)

            {

                scanf("%d",&a[i]);

                scanf("%d",&b[i]);

                scanf("%d",&c[i]);

                scanf("%d",&d[i]);

            }

  

        intcnt=0;

        for(inti=0;i<n;i++)

        {

            for(intj=0;j<n;j++)

            {

                cd[cnt++]=c[i]+d[j];

  

            }

        }

        sort(cd,cd+cnt);

         longlong ans=0;

        for(inti=0;i<n;i++)

        {

  

            for(intj=0;j<n;j++)

            {

                inttmp=0-(a[i]+b[j]);

              //  printf("a:%d  b:%d    num:%d\n",a[i],b[i],upper_bound(cd,cd+cnt,tmp)-lower_bound(cd,cd+cnt,tmp));

                ans+=upper_bound(cd,cd+cnt,tmp)-lower_bound(cd,cd+cnt,tmp);

  

  

            }

  

        }

        printf("%lld\n",ans);

  

  

  

    }

  

    return0;

}