POJ 2785 4 Values whose Sum is 0（折半枚举）

Description

Input

Output

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

思路：先把前两列枚举两两组合产生N*N个数，然后在枚举后两列，然后二分求可以等于0的个数就好 复杂度是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;
}

POJ 2785 4 Values whose Sum is 0（折半枚举）

原文：http://blog.csdn.net/kalilili/article/details/46778727