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| ##河南省2021CCPC 数学期望 (组合数学+乘法逆元)
每个技能性质相同，只需求出一个技能的贡献值*m即可；
枚举技能使用1--n次的贡献值，对于每个贡献值*m
当前技能使用i次，则剩余(m-1)个技能，占用(n-i)个位置，
#pragma GCC optimize(3,"Ofast","inline")
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
typedef long long ll;
typedef long long LL;
#define int long long
#define ld long double
const int N=1e5+10;
int n,a[100010],m;
const int mod=1e9+7;
int exgcd(int a,int b,int &x,int &y){
if (!b) {
x = 1, y = 0;
return a;
}
int d = exgcd(b, a % b, y, x);
y -= a / b * x;
return d;
}
ll inverse(int a,int p,int ok){
int x=0,y=0;
int res=exgcd(a,p,x,y);
if(res==1) res=(x+p)%p;
else{res=-1;}
return res;
}
int fact[N], infact[N];
int qmi(int a, int k, int p)
{
int res = 1;
while (k)
{
if (k & 1) res = (LL)res * a % p;
a = (LL)a * a % p;
k >>= 1;
}
return res;
}
void init(){
fact[0] = infact[0] = 1;
for (int i = 1; i < N; i ++ )
{
fact[i] = (LL)fact[i - 1] * i % mod;
infact[i] = (LL)infact[i - 1] * qmi(i, mod - 2, mod) % mod;
}
}
int c(int a,int b){
return fact[a] * infact[b] % mod * infact[a - b] % mod;
}
void solve()
{
cin>>n>>m;
int res=0;
for(int i=1;i<=n;i++){
res=res+(LL)i*i%mod*m%mod*c(n,i)%mod*qmi(m-1,n-i,mod);
res=res%mod;
}
res=res*inverse(qmi(m,n,mod),mod,0)%mod;
cout<<res<<endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
init();
int t; cin>>t;
while(t--)
{
solve();
}
return 0;
}
|
