E - 4 Values whose Sum is 0

Description
The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have 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 have n lines containing four integer values (with absolute value as large as 2 28 ) that belong respectively to A, B, C and D .

Output
For each input file, your program has 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 -46
26 -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), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

4组数据求和，先两两划分，分别求两组数的和，再对其中一组和数组查重并记录，再枚举另一个和数组的数据，二分搜索查找已经查重数组的数据即可。
代码如下：

#include<iostream>#include<algorithm>#include<cmath>#include<stdio.h>#include<map>#define MAXN 4000+10typedef long long LL;using namespace std;LL a[MAXN];LL b[MAXN];LL c[MAXN];LL d[MAXN];LL ab[MAXN * MAXN];LL cd[MAXN * MAXN];map<LL,LL> ma;int bs(LL key, int n){    int lo = 0, hi = n*n - 1;    int mi;    while(lo <= hi)    {        mi = ((hi-lo)>>1) + lo;        //cout << mi << endl;        //cout << key + cd[mi] << endl;        while(key + cd[mi] == 0)            return mi;        if(key + cd[mi] < 0)            lo = mi + 1;        else            hi = mi - 1;        //cout << lo << " " << hi << endl;    }    return -1;}int main(){    int n;    int sum;    //freopen("in.txt","r",stdin);    while(cin >> n)    {        sum = 0;        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++)                ab[i*n+j] = a[i] + b[j];        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 temp = cd[0];        ma.insert(make_pair(cd[0],1));        for(int i = 1; i < n*n; i++)        {            if(cd[i] == temp)                ma[cd[i]]++;            else            {                temp = cd[i];                ma[temp]++;            }        }        for(int i = 0; i < n*n; i++)        {            //cout << ab[i] << endl;            int mi = bs(ab[i], n);            //cout << ab[i] << " " << cd[0] << endl;            if(mi != -1)            {                //cout << ab[i] << endl;                sum += ma[cd[mi]];               // cout << ma[cd[mi]] << endl;                //cout << ma[cd[mi]] << endl;            }        }        cout << sum << endl;        ma.clear();    }    return 0;}