Day3-J-4 Values whose Sum is 0 POJ2785

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

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).

思路：直接暴力的复杂度是N^4，肯定不行，可以两两合并，复杂度就是N^2，代码如下：

vector<int> buf[];

vector<long long>sum1,sum2;

int main() {

    int N,sum = ;

    scanf("%d",&N);

    for(int i = ; i < N; ++i) { // read

        for(int j = ; j < ; ++j) {

            int tmp;

            scanf("%d",&tmp);

            buf[j].push_back(tmp);

        }

    }

    for(int i = ; i < N; ++i) {

        for(int j = ; j < N; ++j) {

            sum1.push_back(buf[][i] + buf[][j]);

            sum2.push_back(buf[][i] + buf[][j]);

        }

    }

    sort(sum1.begin(), sum1.end());

    sort(sum2.begin(), sum2.end());

    for(int i = ;i < sum1.size(); ++i) {

        long long tmp = -sum1[i];

        sum += upper_bound(sum2.begin(), sum2.end(), tmp) - lower_bound(sum2.begin(), sum2.end(), tmp);

    }

    printf("%d",sum);

    return ;

}

巴特西

Day3-J-4 Values whose Sum is 0 POJ2785

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