NC13884. Paint Box
描述
输入描述
The first line of the input contains integer T(1≤T≤100) -the number of the test casesFor each case: there will be one line, which contains three integers n,m,k(1≤n,m≤109 1≤k≤106, k≤n,m).
输出描述
For each test case, you need print an integer means the number of ways of different coloring methods modulo 109+7.
示例1
输入:
2 3 2 2 3 2 1
输出:
2 0
说明:
In the first example we have two ways:C++(clang++11) 解法, 执行用时: 297ms, 内存消耗: 376K, 提交时间: 2021-02-15 11:33:23
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
ll fast(ll a,ll b)
{
ll ans=1;
a%=mod;
while(b)
{
if(b&1)
ans=(ans*a)%mod;
a=(a*a)%mod;
b>>=1;
}
return ans;
}
ll C(ll n,ll r)
{
if(n<r)
return 0;
if(n-r<r)
r=n-r;
ll a=1,b=1;
for(int i=0;i<r;i++)
a=(a*(n-i))%mod,b=(b*(i+1))%mod;
return a*fast(b,mod-2)%mod;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
ll n,m,k,i,res=1,ans=0;
scanf("%lld %lld %lld",&n,&m,&k);
for(i=k;i>=1;i--)
{
ans=(ans+i*fast(i-1,n-1)%mod*res%mod+mod)%mod;
res=-1*res*i%mod*fast(k-i+1,mod-2)%mod;
}
ans=(ans*C(m,k))%mod;
printf("%lld\n",ans);
}
return 0;
}