Educational Codeforces Round 7 F. The Sum of the k-th Powers

 

重心拉格朗日插值定理可以解决求和公式。。但是我的跑的也太慢了吧。。。

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <set>
#include <map>
#include <list>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>
#include <functional>
#include <stack>
using namespace std;
typedef long long ll;
#define T int t_;Read(t_);while(t_--)
#define dight(chr) (chr>='0'&&chr<='9')
#define alpha(chr) (chr>='a'&&chr<='z')
#define INF (0x3f3f3f3f)
#define maxn (1000005)
#define maxm (10005)
#define mod 1000000007
#define ull unsigned long long
#define repne(x,y,i) for(int i=(x);i<(y);++i)
#define repe(x,y,i) for(int i=(x);i<=(y);++i)
#define repde(x,y,i) for(int i=(x);i>=(y);--i)
#define repdne(x,y,i) for(int i=(x);i>(y);--i)
#define ri register int
inline void Read(int &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if(chr=='-')sign=-1;
    for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;}
inline void Read(ll &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if(chr=='-')sign=-1;
    for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;}
ll quickpow(ll x,ll y){
    ll ans = 1;
    while(y){
        if(y & 1){
            ans = ans * x;
            if(ans >= mod) ans %= mod;
        }
        x = x * x;
        if(x >= mod) x %= mod;
        y >>= 1;
    }
    return ans;
}
int main()
{
    freopen("a.in","r",stdin);
    freopen("b.out","w",stdout);
    ll n,k;
    Read(n),Read(k);
    if(n <= k + 2){
        ll ans = 0;
        repe(1,n,i){
            ans = ans + quickpow(i,k);
            if(ans >= mod) ans %= mod;
        }
        printf("%lld\n",ans);
    }
    else{
        ll ans = 0,y = 1,w = 1,t = 1;
        repe(2,k+2,i){
            w = w * (1 - i);
            if(w < 0) w = (w + (-w / mod + 1) * mod) % mod;
        }
        repe(1,k+2,i){
            ans += quickpow(w,mod - 2) * y % mod * quickpow(n-i,mod-2) % mod;
            if(ans >= mod) ans %= mod;
            w = w * i % mod * quickpow(i - k - 2 + mod,mod-2) % mod;
            y = y + quickpow(i + 1,k);
            if(y >= mod) y %= mod;
        }
        repe(1,k+2,i){
            t = t * (n - i);
            if(t >= mod) t %= mod;
        }
        ans = ans * t;
        if(ans >= mod) ans %= mod;
        printf("%lld\n",ans);
    }
}

 

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