POJ 2785 A Values whose Sum is 0 ( 拆合+排序+二分 )

#include <iostream>
#include <fstream>
#include <algorithm>
using namespace std;

#define MAXN 4010

int A[MAXN];
int B[MAXN];
int C[MAXN];
int D[MAXN];
int CD[MAXN * MAXN];

int main(){

int n;
long long res = 0;

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 ){
CD[i * n + j] = C[i] + D[j];
}
}

sort( CD, CD + n * n );

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 );
}
}

cout << res << endl;

return 0;

}