https://www.luogu.org/problemnew/show/P4948
这篇博客主要目的是存一下的dls的神奇板子,本来应该是推公式或者二分做的
但是dls的插值板子直接写好了这个特殊式子的算法......
#include <bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<int,int>
#define ull unsigned long long
#define all(x) x.begin(),x.end()
#pragma GCC optimize("unroll-loops")
#define inline inline __attribute__( \
(always_inline, __gnu_inline__, __artificial__)) \
__attribute__((optimize("Ofast"))) __attribute__((target("sse"))) \
__attribute__((target("sse2"))) __attribute__((target("mmx")))
#define IO ios::sync_with_stdio(false);
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define for_node(x,i) for(int i=head[x];i;i=e[i].next)
#define show(x) cout<<#x<<"="<<x<<endl
#define showa(a,b) cout<<#a<<'['<<b<<"]="baidu<a[b]<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
using namespace std;
const int maxn=1e6+10,maxm=2e6+10;
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
const double PI=acos(-1.0);
//head
int casn,n,m,k;
int num[maxn];
ll a[maxn];
ll pow_mod(ll a,ll b,ll c=mod,ll ans=1){while(b){if(b&1) ans=(a*ans)%c;a=(a*a)%c,b>>=1;}return ans;} namespace polysum {
const int maxn=101000;
const ll mod=1e9+7;
ll a[maxn],f[maxn],g[maxn],p[maxn],p1[maxn],p2[maxn],b[maxn],h[maxn][2],C[maxn];
ll calcn(int d,ll *a,ll n) {//d次多项式(a[0-d])求第n项
if (n<=d) return a[n];
p1[0]=p2[0]=1;
rep(i,0,d) {
ll t=(n-i+mod)%mod;
p1[i+1]=p1[i]*t%mod;
}
rep(i,0,d) {
ll t=(n-d+i+mod)%mod;
p2[i+1]=p2[i]*t%mod;
}
ll ans=0;
rep(i,0,d) {
ll t=g[i]*g[d-i]%mod*p1[i]%mod*p2[d-i]%mod*a[i]%mod;
if ((d-i)&1) ans=(ans-t+mod)%mod;
else ans=(ans+t)%mod;
}
return ans;
}
void init(int maxm) {//初始化预处理阶乘和逆元(取模乘法)
f[0]=f[1]=g[0]=g[1]=1;
rep(i,2,maxm+4) f[i]=f[i-1]*i%mod;
g[maxm+4]=pow_mod(f[maxm+4],mod-2);
per(i,1,maxm+3) g[i]=g[i+1]*(i+1)%mod;
}
ll polysum(ll n,ll *a,ll m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]
// m次多项式求第n项前缀和
a[m+1]=calcn(m,a,m+1);
rep(i,1,m+1) a[i]=(a[i-1]+a[i])%mod;
return calcn(m+1,a,n-1);
}
ll qpolysum(ll R,ll n,ll *a,ll m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]*R^i
if (R==1) return polysum(n,a,m);
a[m+1]=calcn(m,a,m+1);
ll r=pow_mod(R,mod-2),p3=0,p4=0,c,ans;
h[0][0]=0;
h[0][1]=1;
rep(i,1,m+1) {
h[i][0]=(h[i-1][0]+a[i-1])*r%mod;
h[i][1]=h[i-1][1]*r%mod;
}
rep(i,0,m+1) {
ll t=g[i]*g[m+1-i]%mod;
if (i&1) p3=((p3-h[i][0]*t)%mod+mod)%mod,p4=((p4-h[i][1]*t)%mod+mod)%mod;
else p3=(p3+h[i][0]*t)%mod,p4=(p4+h[i][1]*t)%mod;
}
c=pow_mod(p4,mod-2)*(mod-p3)%mod;
rep(i,0,m+1) h[i][0]=(h[i][0]+h[i][1]*c)%mod;
rep(i,0,m+1) C[i]=h[i][0];
ans=(calcn(m,C,n)*pow_mod(R,n)-c)%mod;
if (ans<0) ans+=mod;
return ans;
}
} int main() {
//#define test
#ifdef test
auto _start = chrono::high_resolution_clock::now();
freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#endif
IO;
ll n,r,k;
cin>>n>>r>>k;
polysum::init(k+5);
rep(i,0,2010) a[i]=pow_mod(i,k);
ll ans=polysum::qpolysum(r,n+1,a,k+1);
if(k==0) ans=(ans-1+mod)%mod;
cout<<ans<<endl;
#ifdef test
auto _end = chrono::high_resolution_clock::now();
cerr << "elapsed time: " << chrono::duration<double, milli>(_end - _start).count() << " ms\n";
fclose(stdin);fclose(stdout);system("out.txt");
#endif
return 0;
}