UVa 1152 4Values whose Sum is 0

題意 從4個n元集中各挑出一個數 使它們的和為零有多少種方法

直接n^4枚舉肯定會超時的 可以把兩個集合的元素和放在數組裡 然後排序 枚舉另外兩個集合中兩元素和 看數組中是否有其相反數就行了 復雜度為n^2*logn

#include 
#define l(i) lower_bound(s,s+m,i)
#define u(i) upper_bound(s,s+m,i)
using namespace std;
const int N = 4005;
int a[N], b[N], c[N], d[N], s[N * N];

int main()
{
    int cas, m, k, n;
    long long ans;
    scanf("%d", &cas);
    while(cas--)
    {
        ans = m = 0;
        scanf("%d", &n);
        for(int i = 0; i < n; ++i)
            scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j)
                s[m++] = a[i] + b[j];
        sort(s, s + m);
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j)
                k = -c[i] - d[j], ans += (u(k) - l(k));

        printf("%lld\n", ans);
        if(cas) puts("");
    }
    return 0;
}

1152 - 4 Values whose Sum is 0

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 input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by 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 (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 2²⁸ ) 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 consecutive cases will be separated by a blank line.

For each input file, your program has to write the number quadruplets whose sum is 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 is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).