UVa 1152 和为0的4个值（二分查找）

https://vjudge.net/problem/UVA-1152

题意：给定4个n元素集合A,B,C,D，要求分别从中选取一个元素a,b,c,d，使得a+b+c+d=0。问有多少种取法。

思路：直接暴力枚举的话是会超时的。可以选把a+b的值枚举出来存储，c和d的值也一样并排序，这样就可以在c和d中进行二分查找了。

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

const int maxn = 4000 + 5;

int n;
int a[maxn], b[maxn], c[maxn], d[maxn];
int s1[25000000], s2[25000000];


void solve()
{
    int k = 0;
    int cnt = 0;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            s1[k] = a[i] + b[j];
            s2[k] = c[i] + d[j];
            k++;
        }
    }
    sort(s2, s2 + k);
    for (int i = 0; i < k; i++)
    {
        int t = s1[i];
        int left = 0, right = k-1;
        while (left <= right)
        {
            int mid = (left + right) / 2;
            if (s2[mid] == -t)
            {
                cnt++;
                int p1 = mid;
                int p2 = mid;
                //因为有可能存在一样的数，所以前后还需要判断
                while (s2[++p1] == -t && p1<k )  cnt++;
                while (s2[--p2] == -t && p2>=0)  cnt++;
                break;
            }
            else if (s2[mid]>-t)  right = mid-1;
            else left = mid+1;
        }
    }
    cout << cnt << endl;
}

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    int t;
    cin >> t;
    while (t--)
    {
        cin >> n;
        for (int i = 0; i < n; i++)
        {
            cin >> a[i] >> b[i] >> c[i] >> d[i];
        }
        solve();
        if (t)  cout << endl;
    }
    return 0;
}