4 Values whose Sum is 0 --CSU-ACM2017暑假集训2-二分搜索

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).

题目意思是只要4个数只要来自4个数组的不同位置且相加为0就算一组，两个4元序列完全相等也算新的一组。

另，注释的地方跪了好多遍。

AC代码：

#include<iostream>#include<cstdio>#include<algorithm>#include<vector>using namespace std;int main(){    long long n;    while(~scanf("%lld",&n))    {        vector<long long> a,b,c,d,sum1,sum2;        for(long long i=1;i<=n;i++)        {            long long x1,x2,x3,x4;            scanf("%lld%lld%lld%lld",&x1,&x2,&x3,&x4);            a.push_back(x1);            b.push_back(x2);            c.push_back(x3);            d.push_back(x4);        }        for(long long i=0;i<n;i++)            for(long long j=0;j<n;j++)            sum1.push_back(a[i]+b[j]);        for(long long i=0;i<n;i++)            for(long long j=0;j<n;j++)            sum2.push_back(c[i]+d[j]);        sort(sum1.begin(),sum1.end());        sort(sum2.begin(),sum2.end());        long long cnt=0;        for(long long i=0;i<n*n;i++)//之前写成i=1;i<=n*n;而WA了无数遍。        {            long long left=lower_bound(sum2.begin(),sum2.end(),0-sum1[i])-sum2.begin();            long long right=upper_bound(sum2.begin(),sum2.end(),0-sum1[i])-sum2.begin();            cnt+=(long long)(right-left);    //        if(right-left!=0)cout<<sum1[i]<<endl;        }        cout<<cnt<<endl;    }}