Description
求组合数C(n,m)
Input
T（0 < T < 2000)
每组样例两个数n, m（0 < n, m< 60）
Output
C(n,m)
Sample Input
2
1 1
2 1
Sample Output
1
2
Source
Unknown
思路：真是一道神奇的题啊，我想着用常规的办法去求解，但是爆了long long也不自知，真是傻。然后用了优化。
#include <cstdio>
#include <iostream>
#include <cmath>
#include <
string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define ll long long
int t, n, m;
ll sum1;
int main()
{
scanf("%d", &
t);
while(t--
)
{
sum1 =
1;
scanf("%d%d", &n, &
m);
for(
int i =
1; i <= m; i++
)
{
sum1 *= (n-i+
1);
sum1 /=
i;
}
printf("%lld\n", sum1);
}
return 0;
}
转载于:https://www.cnblogs.com/RootVount/p/10991748.html
