poj 2785 4 Values whose Sum is 0

4 Values whose Sum is 0 Time Limit: 15000MS Memory Limit: 228000KTotal Submissions: 10535 Accepted: 2904Case 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 2 ²⁸ ) 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 -46 26 -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). #include <stdio.h>#include <algorithm>using namespace std;const int N=4000;int a[N],b[N],c[N],d[N],ab[N*N],cd[N*N];int main(){int n,pab,pcd,ans,i,j,k;while(scanf("%d",&n)!=EOF) { pab=pcd=ans=0;for(i=0;i<n;i++) scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);for(i=0;i<n;i++)for(j=0;j<n;j++) { ab[pab++]=a[i]+b[j]; cd[pcd++]=-c[i]-d[j]; } sort(cd,cd+pcd);for(i=0;i<pab;i++) {int l=0,r=pcd-1,mid;while(l<=r) { mid=(l+r)/2;if(ab[i]==cd[mid]) { ans++;for(k=mid+1;k<pcd;k++)if(ab[i]==cd[k]) ans++;else break;for(k=mid-1;k>=0;k--)if(ab[i]==cd[k]) ans++;else break;break; }else if(ab[i]<cd[mid]) r=mid-1;else l=mid+1; } } printf("%d\n",ans); } }

转载于:https://www.cnblogs.com/w0w0/archive/2011/11/22/2258655.html