이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
#pragma GCC optimize("unroll-loops,no-stack-protector")
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;
const ll MOD=1e9+7;
const ll MOD2=998244353;
const ll N=4e5+5;
const ll K=350;
const ld pi=3.14159265359;
const ll INF=(1LL<<40);
#define SQ(i) ((i)*(i))
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define setp setprecision
#define lwb lower_bound
#define SZ(_a) (ll)_a.size()
ll n,m,fac[N],rfac[N];
ll C(ll a,ll b){
return fac[a]*rfac[b]%MOD*rfac[a-b]%MOD;
}
ll po(ll base,ll ti){
ll res=1;
for(;ti;ti>>=1,base=base*base%MOD){
if(ti&1)res=res*base%MOD;
}
return res;
}
ll cal(ll k){
ll tmp=0;
for(int i=0,o=1;i<k;i++,o=-o)tmp=(tmp+1LL*o*C(n,i)*po(k-i,n))%MOD;
return (tmp+MOD)%MOD;
}
int main(){
ios_base::sync_with_stdio(0);cin.tie(0);
cin>>n>>m;
fac[0]=rfac[0]=1;
REP1(i,N-1)fac[i]=i*fac[i-1]%MOD,rfac[i]=rfac[i-1]*po(i,MOD-2)%MOD;
ll ans=cal(m)-cal(m-1);
cout<<(ans%MOD+MOD)%MOD<<"\n";
return 0;
}
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