题解 | #纳米兔子#

//标程 注意一下读入就好了（数据还是大，刚才测了两下cin过还是有点费劲，这份是能过的）
#include <iostream>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    long long n, m;
    int a[10][6] = {0};
    a[1][1] = 1;
    a[2][1] = 1,  a[2][2] = 1;
    a[3][1] = 1,  a[3][2] = 2,  a[3][3] = 1;
    a[4][1] = 1,  a[4][2] = 3,  a[4][3] = 3, a[4][4] = 1;
    a[5][1] = 1,  a[5][2] = 4,  a[5][3] = 6, a[5][4] = 4;
    a[6][1] = 5,  a[6][2] = 10, a[6][3] = 10;
    a[7][1] = 15, a[7][2] = 20,
    a[8][1] = 35;
    while(cin >> n >> m)
    {
        long long ans = 0;
        for(int i = 0; i < 5; i++)
        {
            ans += a[m][i];
        }
        cout << ans * n<< '\n';
    }
    return 0;
}