1152 - 4 Values whose Sum is 0

The SUM problemcan be formulated as follows: given four lists A, B, C, D ofinteger values, compute how many quadruplet (a, b, c, d )  A x B x C x D aresuch that a + b + c + d =0 . In the following, we assume that all lists have the same size n.

Input 

The inputbegins with a single positive integer on a line by itself indicating the numberof the cases following, each of them as described below. This line is followedby a blank line, and there is also a blank line between two consecutive inputs.

The first line of the input file contains the size of the lists n (thisvalue can be as large as 4000). We then have n linescontaining four integer values (with absolute value as large as2²⁸ )that belong respectively to A, B, C and D .

Output 

For each test case,the output must follow the description below. The outputs of two consecutivecases will be separated by a blank line.

For each input file, your program has to write the number quadruplets whose sumis zero.

Sample Input 

1

 

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

Sample Explanation: Indeed, the sum of the five following quadruplets iszero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75,77), (-32, -54, 56, 30).

代码：

#include<iostream>

#include<algorithm>

using namespacestd;

 

const int maxn =4000 + 5;

int n, c, A[maxn],B[maxn], C[maxn], D[maxn], sums[maxn*maxn];

 

int main()

{

    ios::sync_with_stdio(false);

    int T;

    cin>>T;

    while(T--)

    {

        cin>>n;

        for(int i = 0; i < n; i++)

        {

           cin>>A[i]>>B[i]>>C[i]>>D[i];

        }

        c = 0;

        for(int i = 0; i < n; i++)

        {

            for(int j = 0; j < n; j++)

            {

                sums[c++] = A[i] + B[j];

            }

        }

        sort(sums, sums+c);

        long long cnt = 0;

        for(int i = 0; i < n; i++)

        {

            for(int j = 0; j < n; j++)

            {

                cnt += upper_bound(sums,sums+c, -C[i]-D[j]) - lower_bound(sums, sums+c, -C[i]-D[j]);

            }

        }

        cout<<cnt<<endl;

        if(T)

        {

            cout<<"\n";

        }

    }

    return 0;

}

解析：

为了防止超时：枚举A+B，并将结果存储在一个数组中，再枚举-C-D，在排序后的A+B数组中进行查找，求的结果。