题目描述
三倍经验题。
给你\(n,m\)，求
\[\sum_{i=1}^ni^mm^i
\]
\]
\(n\leq {10}^9,1\leq m\leq 500000\)
题解
当\(m=1\)时\(ans=\frac{n(n+1)}{2}\)
剩下的部分这篇博客有讲YWW's Blog
时间复杂度：\(O(m+\log n)\)
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll p=1000000007;
ll fp(ll a,ll b)
{
ll s=1;
for(;b;b>>=1,a=a*a%p)
if(b&1)
s=s*a%p;
return s;
}
int pri[100010];
int b[1000010];
int cnt;
ll s[1000010];
ll fac[1000010];
ll ifac[1000010];
ll inv[1000010];
ll f1[1000010];
ll f2[1000010];
ll f[1000010];
ll pre[1000010];
ll suf[1000010];
ll getc(int x,int y)
{
return fac[x]*ifac[y]%p*ifac[x-y]%p;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("bzoj4126.in","r",stdin);
freopen("bzoj4126.out","w",stdout);
#endif
int n,m;
scanf("%d%d",&n,&m);
if(m==1)
{
printf("%lld\n",ll(n)*(n+1)/2%p);
return 0;
}
n++;
fac[0]=fac[1]=ifac[0]=ifac[1]=inv[1]=1;
for(int i=2;i<=m+2;i++)
{
inv[i]=-p/i*inv[p%i]%p;
ifac[i]=ifac[i-1]*inv[i]%p;
fac[i]=fac[i-1]*i%p;
}
s[0]=0;
s[1]=1;
for(int i=2;i<=m+2;i++)
{
if(!b[i])
{
s[i]=fp(i,m);
pri[++cnt]=i;
}
for(int j=1;j<=cnt&&i*pri[j]<=m+2;j++)
{
b[i*pri[j]]=1;
s[i*pri[j]]=s[i]*s[pri[j]]%p;
if(i%pri[j]==0)
break;
}
}
f1[0]=1;
f2[0]=0;
ll invm=fp(m,p-2);
for(int i=1;i<=m+1;i++)
{
f1[i]=f1[i-1]*invm%p;
f2[i]=(f2[i-1]+s[i-1])*invm%p;
}
ll v1=0,v2=0;
for(int i=0;i<=m+1;i++)
{
v1=(v1+((m+1-i)&1?-1:1)*getc(m+1,i)*f1[i])%p;
v2=(v2+((m+1-i)&1?-1:1)*getc(m+1,i)*f2[i])%p;
}
f[0]=-v2*fp(v1,p-2)%p;
for(int i=1;i<=m+1;i++)
f[i]=(f1[i]*f[0]+f2[i])%p;
if(n<=m+1)
{
ll ans=fp(m,n)*f[n]-f[0];
ans=(ans%p+p)%p;
printf("%lld\n",ans);
return 0;
}
for(int i=0;i<=m;i++)
{
pre[i]=n-i;
if(i)
pre[i]=pre[i-1]*pre[i]%p;
}
for(int i=m;i>=0;i--)
{
suf[i]=n-i;
if(i!=m)
suf[i]=suf[i+1]*suf[i]%p;
}
ll ans=0;
for(int i=0;i<=m;i++)
{
ll v=1;
if(i)
v=v*pre[i-1]%p;
if(i!=m)
v=v*suf[i+1]%p;
ans=(ans+f[i]*v%p*ifac[i]%p*ifac[m-i]%p*((m-i)&1?-1:1))%p;
}
ans=fp(m,n)*ans-f[0];
ans=(ans%p+p)%p;
printf("%lld\n",ans);
return 0;
}