POJ 2785 4 Values whose Sum is 0 (binary enumeration)

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 17088		Accepted: 4998
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

Idea: first two columns of enumeration 22 to produce the number of n*n, and then two columns after enumeration, and then two points can be equal to 0 of the number of good complexity is O (N*NLOGN)

#include <iostream> #include <cstring> #include <cmath> #include <queue> #include <stack> #include <list> #include <map> #include <set> #include <sstream> #include <string># Include <vector> #include <cstdio> #include <ctime> #include <bitset> #include <algorithm > #define SZ (x) ((int) (x). Size ()) #define ALL (v) (v). Begin (), (v). End () #define foreach (I, v) for (__typeof ((v). Begin () ) i = (v). Begin (); I! = (v). End (); + + i) #define Refeach (i, v) for (__typeof ((v). Rbegin ()) i = (v). Rbegin (); I! = (v). rend (); + + i) #define REP (i,n) for (int I=1; I<=int (n);    i++) using namespace std;typedef long long ll;const int N = 4000+100;int a[5][n];int all[n*n];int half[n*n];int main () {    int n;        while (~SCANF ("%d", &n)) {Rep (i,n) Rep (j,4) scanf ("%d", &a[j][i]);        int top = 0;        ll ans = 0;        Rep (I,n) Rep (j,n) half[++top] = A[1][i]+a[2][j];    Rep (I,n) Rep (j,n) all[top--] = A[3][i]+a[4][j];    Sort (all+1,all+n*n+1);        REP (i,n*n) ans + = Upper_bound (All+1,all+n*n+1,-half[i])-lower_bound (All+1,all+n*n+1,-half[i]);    printf ("%i64d\n", ans); } return 0;}

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POJ 2785 4 Values whose Sum is 0 (binary enumeration)