[容斥] 51Nod1829 函数

简单的容斥

∑i=0m(−1)i(m−i)n(mi)

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=1000005,MOD=1e9+7;typedef long long LL;int n,m,a[maxn],ans;LL fac[maxn],inv[maxn],fac_inv[maxn];LL C(int n,int m){ return fac[n]*fac_inv[m]%MOD*fac_inv[n-m]%MOD; }LL Pow(LL a,int b){    LL res=1;    for(;b;b>>=1,a=a*a%MOD) if(b&1) res=(res*a)%MOD;    return res;}int main(){    //freopen("A.in","r",stdin);    //freopen("A.out","w",stdout);    fac[0]=1; for(int i=1;i<=1000000;i++) fac[i]=fac[i-1]*i%MOD;    inv[1]=1; for(int i=2;i<=1000000;i++) inv[i]=(LL)(MOD-MOD/i)*inv[MOD%i]%MOD;    fac_inv[0]=1; for(int i=1;i<=1000000;i++) fac_inv[i]=fac_inv[i-1]*inv[i]%MOD;    scanf("%d%d",&n,&m);    for(int i=0;i<=m;i++)     (ans+=Pow(m-i,n)*C(m,i)*(i&1?-1:1)%MOD)%=MOD;    printf("%d\n",(ans+MOD)%MOD);    return 0;}