poj2785(4 Values whose Sum is 0)

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4 Values whose Sum is 0

Time Limit: 15000MS Memory Limit: 228000KTotal Submissions: 19972 Accepted: 5953Case 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

题目意思很简单明了，就是从四个数列中各取一个数，使四个数的和为0，求共有多少种方法？

可以直接的四个for循环，但是显然很不高效。把它们分成AB和CD再考虑，就可以快速求解。大致思路是把C、D中的数字枚举出来，排好序，为了使总和为0则需要从C、D中取出c+d=-a-b，可以用二分查找，时间复杂度为O(n^2logn)。

AC代码:

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;#define maxn 4005int A[maxn],B[maxn],C[maxn],D[maxn];int CD[maxn*maxn];int main(){    int n;    scanf("%d",&n);    for(int i=0;i<n;i++){        scanf("%d%d%d%d",&A[i],&B[i],&C[i],&D[i]);    }    for(int i=0;i<n;i++){        for(int j=0;j<n;j++){            CD[i*n+j]=C[i]+D[j];        }    }    sort(CD,CD+n*n);    long long res=0;    for(int i=0;i<n;i++){        for(int j=0;j<n;j++){            int cd=-(A[i]+B[j]);            res+=upper_bound(CD,CD+n*n,cd)-lower_bound(CD,CD+n*n,cd);        }    }    printf("%lld\n",res);    return 0;}