Time Limit: 15000MS | Memory Limit: 228000K | |
Total Submissions: 30579 | Accepted: 9351 | |
Case 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 -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).
Source
#include<iostream> #include<algorithm> using namespace std; typedef long long LL; const int maxn = 4010; int n; LL A[maxn],B[maxn],C[maxn],D[maxn]; LL AB[maxn*maxn],CD[maxn*maxn]; int main(void) { cin >> n; for(int i=0;i<n;i++) { cin >> A[i]>>B[i]>>C[i]>>D[i]; } for(int i=0;i<n;i++) { for(int j=0;j<n;j++) { AB[i*n+j]=A[i]+B[j]; CD[i*n+j]=C[i]+D[j]; } } sort(CD,CD+n*n); int result = 0; for(int i=0;i<n*n;i++) { result += upper_bound(CD,CD+n*n,-AB[i])-lower_bound(CD,CD+n*n,-AB[i]); //cout << result<<endl; } cout << result; return 0; }