題目鏈接:http://swjtuoj.cn/problem/2382/
C(n,m) = (n*(n-1)*……*(n-m+1)) / m! = (n/1) * ((n-1)/2) * ((n-2) / 3) * …… * ((n-m+1) / 1)
除法用乘法逆元
具體見代碼:
#include <iostream>
#include <cstdio>
#include <cmath>
#include <map>
using namespace std;
typedef long long ull;
const ull mod = 1e9 + 7;
ull ksp(ull a, ull b)//知識點1,快速冪求乘法逆元
{
ull ret = 1;
while (b > 0)
{
if (b % 2 == 1)
ret = (ret * a) % mod;
b >>= 1;
a = (a * a) % mod;
}
return ret;
}
ull C(ull n, ull m)//知識點2
{
ull ret = 1;
for (int i=1; i<=m; i++)
{
ret = ret * (n-i+1) % mod * ksp(i, mod-2) % mod;// (n-i+1)/i == (n-i+1)*(i 的逆元)
}
return ret;
}
int main()
{
int t;
cin >> t;
for (int _=0; _<t; _++)
{
ull n, m, k;
scanf("%lld%lld%lld", &n, &m, &k);
ull ans = 0;
ull cmk = 1;
for (int i=1; i<=k; i++)
{
cmk = (((cmk * (k-i+1)) % mod) * ksp(i, mod-2)) % mod;
ull ret = (((cmk * i) % mod) * ksp(i-1, n-1)) % mod;
if ((k-i) % 2 == 1)
ans = (ans + mod - ret) % mod;
else
ans = (ans + ret) % mod;
}
ans = (ans * C(m,k)) % mod;
printf("%lld\n", ans);
}
return 0;
}