POJ 2785 4 Values whose Sum is 0

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 20015		Accepted: 5974
Case Time Limit: 5000MS

Description

The SUM problem can formulated as Follows:given four lists A, B, C, D of an integer values, compute how many quadruplet ( A, B, C, D) ∈a x B x C x D is such that A + B + c + d = 0. In the following, we assume this all lists has 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 had 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 of the input file, your program have 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), (-3) 2, 22, 56,-46), (-32, 30,-75, 77), (-32,-54, 56, 30).

Source

Southwestern Europe 2005    parsing: Binary enumeration. directly from the 4 series selected words have N4, time complexity is too high, not advisable. Binary enumeration can be taken, a, B, C, D into AB and CD considerations, from A, b after a, b, in order to make the sum of 0 need to remove c+d =-(a+b) from C, D. So first the C, D take the number of N2 method enumeration out, side-by good order, so you can use binary search, time complexity of O (N2LOGN2).    

#include <cstdio> #include <algorithm> #include <vector> #define LL long longusing namespace Std;const int MAXN = 4000+5;int A[MAXN], B[MAXN], C[MAXN], D[maxn];int cd[maxn*maxn];int n;pair<vector<int>::iterator, Vector<int>::iterator> it;void Solve () {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);    ll res = 0;    for (int i = 0, i < n; ++i) {for        (int j = 0; J < N; ++j) {            int c_d =-(A[i]+b[j]);            it = Equal_range (CD, Cd+n*n, c_d);            Res + = It.second-it.first;        }    }    printf ("%i64d\n", res);} int main () {    scanf ("%d", &n);    for (int i = 0; i < n; ++i)        scanf ("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);    Solve ();    return 0;}
   

　　

　　

POJ 2785 4 Values whose Sum is 0