【BZOJ】【3157】&【BZOJ】【3516】国王奇遇记

数论

　　题解：http://www.cnblogs.com/zhuohan123/p/3726933.html

　　copy一下推导过程：

令$$S_i=sum_{k=1}^{n}k^im^k$$

我们有$$ egin{aligned} (m-1)S_i &= mS_i-S_i \&=sum_{k=1}^n k^im^{k+1}-sum_{k=1}^n k^i m^k \&=sum_{k=2}^{n+1}(k-1)^i m^k-sum_{k=1}^n k^i m^k \&=n^i m^{n+1}+sum_{k=1}^n m^k ( (k-1)^i-k^i ) \&=n^i m^{n+1}+sum_{k=1}^n ig( sum_{j=1}^{i-1}(-1)^{i-j} inom{i}{j}k^jm^kig) \&=n^i m^{n+1}+sum_{j=0}^{i-1}(-1)^{i-j}inom{i}{j} S_j end{aligned} $$

 1 /**************************************************************
 2     Problem: 3516
 3     User: Tunix
 4     Language: C++
 5     Result: Accepted
 6     Time:372 ms
 7     Memory:1300 kb
 8 ****************************************************************/
 9  
10 //BZOJ 3157
11 #include<cstdio>
12 #include<cstring>
13 #include<cstdlib>
14 #include<iostream>
15 #include<algorithm>
16 #define rep(i,n) for(int i=0;i<n;++i)
17 #define F(i,j,n) for(int i=j;i<=n;++i)
18 #define D(i,j,n) for(int i=j;i>=n;--i)
19 #define pb push_back
20 using namespace std;
21 typedef long long LL;
22 const int N=1010;
23 const LL P=1000000007;
24 /*******************template********************/
25 #define sqr(x) (x)*(x)
26 LL n,m;
27 LL fac[N],inv[N],s[N];
28 LL C(int a,int b){return fac[a]*inv[b]%P*inv[a-b]%P;}
29 LL Pow(LL a,LL b){
30     LL r=1;
31     for(;b;b>>=1,a=a*a%P) if (b&1) r=r*a%P;
32     return r;
33 }
34 int main(){
35 #ifndef ONLINE_JUDGE
36     freopen("3157.in","r",stdin);
37     freopen("3157.out","w",stdout);
38 #endif
39     scanf("%lld%lld",&n,&m);
40     if (m==1){ printf("%lld
",(n+1)*n/2%P);return 0;}
41     fac[0]=1; F(i,1,m) fac[i]=fac[i-1]*i%P;
42     inv[m]=Pow(fac[m],P-2);
43     inv[0]=1;
44     D(i,m-1,1) inv[i]=inv[i+1]*(i+1)%P;
45     s[0]=((Pow(m,n+1)-m)%P+P)%P*Pow(m-1,P-2)%P;
46     F(i,1,m){
47         s[i]=Pow(n,i)*Pow(m,n+1)%P;
48         rep(j,i) s[i]=((s[i]+((i-j)%2==1 ? -1 : 1)*C(i,j)*s[j])%P+P)%P;
49         s[i]=s[i]*Pow(m-1,P-2)%P;
50     }
51     printf("%lld
",s[m]);
52     return 0;
53 }

View Code

3157: 国王奇遇记

Time Limit: 10 Sec Memory Limit: 512 MB
Submit: 357 Solved: 196
[Submit][Status][Discuss]

Description

http://www.lydsy.com/JudgeOnline/upload/201305/1(2).jpg

Input

共一行包括两个正整数N和M。

Output

共一行为所求表达式的值对10^9+7取模的值。

Sample Input

5 3

Sample Output

36363

HINT

1<=N<=10^9,1<=M<=200

Source

Katharon+#1

[Submit][Status][Discuss]