hdu 2837 坑题。

Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1414    Accepted Submission(s): 291


Problem Description
Assume that f(0) = 1 and 0^0=1. f(n) = (n%10)^f(n/10) for all n bigger than zero. Please calculate f(n)%m. (2 ≤ n , m ≤ 10^9, x^y means the y th power of x).
 

 

Input
The first line contains a single positive integer T. which is the number of test cases. T lines follows.Each case consists of one line containing two positive integers n and m.
 

 

Output
One integer indicating the value of f(n)%m.
 

 

Sample Input
2 24 20 25 20
 

 

Sample Output
16 5
 

 

Source
2009 Multi-University Training Contest 3 - Host by WHU
 
 1 /**
 2 a ^ b % c= a ^ ( b % phi ( c )  +  phi ( c ) ) % c      条件:（b>=c）  phi(n)为n的欧拉函数值
 3 我想说(⊙o⊙)…，完全搞不懂标程
 4 **/
 5 #include<iostream>
 6 #include<stdio.h>
 7 #include<cstring>
 8 #include<cstdlib>
 9 using namespace std;
10 typedef __int64 LL;
11 
12 LL Euler(LL n)
13 {
14     LL temp =n,i;
15     for(i=2;i*i<=n;i++)
16     {
17         if(n%i==0)
18         {
19             while(n%i==0)
20                 n=n/i;
21             temp=temp/i*(i-1);
22         }
23     }
24     if(n!=1) temp=temp/n*(n-1);
25     return temp;
26 }
27 LL pow_mod(LL a,LL b,LL p){
28     LL ans=1;
29     while(b)
30     {
31         if(b&1) ans=(ans*a)%p;
32         b=b>>1;
33         a=(a*a)%p;
34     }
35     return ans;
36 }
37 LL fuck(LL a,LL n,LL m)
38 {
39     LL i,ans=1;
40     for(i=1;i<=n;i++)
41     {
42         ans=ans*a;
43         if(ans>=m) return ans;
44     }
45     return ans;
46 }
47 LL dfs(LL n,LL m){
48    LL ans,p,nima;
49    if(n==0) return 1;
50    if(n<10) return n;
51    p = Euler(m);
52    ans=dfs(n/10,p);
53 
54    nima = fuck(n%10,ans,m);
55    if(nima>=m){
56         ans=pow_mod(n%10,ans%p+p,m);
57         if(ans==0) return m;
58    }
59    else{
60        ans=pow_mod(n%10,ans,m);
61    }
62    return ans;
63 }
64 int main()
65 {
66     int T;
67     LL n,m;
68     scanf("%d",&T);
69     while(T--){
70         scanf("%I64d%I64d",&n,&m);
71         printf("%I64d\n",dfs(n,m)%m);
72     }
73     return 0;
74 }

 

posted @ 2014-05-07 23:07  芷水  阅读(408)  评论(0)  编辑  收藏  举报