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 }