HDU/HDOJ 3923 2011 BJTU多校联合 波利亚原理

Invoker

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 122768/62768 K (Java/Others)
Total Submission(s): 515Accepted Submission(s): 163

Problem Description

On of Vance's favourite hero is Invoker, Kael. As many people knows Kael can control the elements and combine them to invoke a powerful skill. Vance like Kael very much so he changes the map to make Kael more powerful.

In his new map, Kael can control n kind of elements and he can put m elements equal-spacedly on a magic ring and combine them to invoke a new skill. But if a arrangement can change into another by rotate the magic ring or reverse the ring along the axis, they will invoke the same skill. Now give you n and m how many different skill can Kael invoke? As the number maybe too large, just output the answer mod 1000000007.

Input

The first line contains a single positive integer T( T <= 500 ), indicates the number of test cases.
For each test case: give you two positive integers n and m. ( 1 <= n, m <= 10000 )

Output

For each test case: output the case number as shown and then output the answer mod 1000000007 in a line. Look sample for more information.

Sample Input

2 3 4 1 2

Sample Output

Case #1: 21 Case #2: 1

Hint

For Case #1: we assume a,b,c are the 3 kinds of elements. Here are the 21 different arrangements to invoke the skills / aaaa / aaab / aaac / aabb / aabc / aacc / abab / / abac / abbb / abbc / abcb / abcc / acac / acbc / / accc / bbbb / bbbc / bbcc / bcbc / bccc / cccc /

Source

2011 Multi-University Training Contest 9 - Host by BJTU

这个题是一个很裸露的波利亚定理的运用

但是最后取余要用逆元。

但是这个题比较巧妙，因为mod=1000000007，他是一个质数，于是要求某一个数的逆元的话直接mod-2次方，然后二分幂去模就行了

至于波利亚定理，我同样使用的POJ 2154以及POJ 2409的写法

我的代码：

#include<stdio.h> #include<string.h> typedef __int64 ll; const ll mod=1000000007; ll eular(ll n) { ll ret=1,i; for(i=2;i*i<=n;i++) { if(n%i==0) { ret=ret*(i-1); n=n/i; while(n%i==0) { n=n/i; ret=ret*i; } } if(n==1) break; } if(n>1) ret=ret*(n-1); return ret%mod; } ll exmod(ll p,ll n) { ll sq=1; while(n>0) { if(n%2==1) sq=(sq%mod)*(p%mod)%mod; p=(p%mod)*(p%mod)%mod; n=n/2; } return sq%mod; } int main() { ll n,ans,i,m,t,T; scanf("%I64d",&T); for(t=1;t<=T;t++) { scanf("%I64d%I64d",&m,&n); ans=0; for(i=1;i*i<n;i++) { if(n%i==0) { ans=(ans+eular(n/i)*exmod(m,i))%mod; ans=(ans+eular(i)*exmod(m,n/i))%mod; } } if(i*i==n) ans=(ans+eular(n/i)*exmod(m,i))%mod; if(n&1) ans=(ans+exmod(m,n/2+1)*n)%mod; else ans=(ans+exmod(m,n/2)*n/2+exmod(m,n/2+1)*n/2)%mod; ll re=exmod(2*n,mod-2); printf("Case #%I64d: %I64d\n",t,(ans%mod)*(re%mod)%mod); } return 0; }