There are well-known formulas: ,
,
. Also mathematicians found similar formulas for higher degrees.
Find the value of the sum modulo 109 + 7 (so you should find the remainder after dividing the answer by the value 109 + 7).
1 #include"bits/stdc++.h"
2 using namespace std;
3
4 #define int long long
5
6 int n,k;
7 const int mod = 1e9+7;
8 const int N = 2e6;
9 int fac[N];
10
11 int ksm(int a,int b)
12 {
13 int ans = 1;
14 a%=mod;
15 for(; b; b>>=1,a*=a,a%=mod)
16 if(b&1)ans*=a,ans%=mod;
17 return ans;
18 }
19 void lg()
20 {
21
22 int ans = 0;
23 fac[1]=1; fac[0]=1;
24 for(int i=2; i<=k+10; i++)fac[i]=(fac[i-1]*i%mod);
25 ///Pretreatment factorial
26 is int now = 0 ;
27 int FZ = . 1 ;
28 for ( int I = . 1 ; I <= K + 2 ; I ++ )
29 FZ * = (Ni), FZ = MOD%; /// Processing molecules
30
31 is for ( int I = . 1 ; I <= K + 2 ; I ++ )
32 {
33 is now KSM + = (I, K); /// each of Yi
34 is now% = MOD;
35 int INV1 KSM = (Ni, mod- 2); /// small molecule inverse element
36 int INV2 = KSM (FAC [I- . 1 ] * FAC [K + 2 -i], mod- 2 );
37 [ /// denominator inverse of the denominator can be seen is the product of two factorial
38 is
39 int F = . 1 ;
40 IF ((K + 2 -i)% 2 == . 1 ) F = - . 1 ;
41 is ANS + FZ = F * * * INV1 INV2% MOD% MOD% * now MOD;
42 is ANS% = MOD;
43 is
44 is // COUT I << << "" << endl << ANS;
45 }
46 is COUT << (ANS + MOD)%mod;
47
48
49 }
50
51 signed main()
52 {
53 //cout<<ksm(2,5);
54 cin>>n>>k;
55 if(k==0)
56 {
57 cout<<n;
58 return 0;
59 }
60 if(n<=k+2)
61 {
62 int now=0;
63 for(int i=1; i<=n; i++)
64 now += ksm(i,k),now%=mod;
65 cout<<now;
66 return 0;
67
68 }
69 lg();
70
71 }
Input
The only line contains two integers n, k (1 ≤ n ≤ 109, 0 ≤ k ≤ 106).
OutputPrint the only integer a — the remainder after dividing the value of the sum by the value 109 + 7.
Examples
Input
4 1
Output
10
Input
4 2
Output
30
Input
4 3
Output
100
Input
4 0
Output
4
When the board took