hdu 2837 Calculation 指数循环节套路题

Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2272    Accepted Submission(s): 536


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
 

题意：求解$f(n) = {(n\%10)}^{f(n/10)}\%m $，其中$(2 <= n, m <= 10^9)$;

思路：一看到指数模除，并且没有说明是否互素，直接上指数循环节，化简指数==套路...

注：

1. 使用指数循环节都是递归形式，所以欧拉函数也需要是递归形式；

2. 在指数循环节中的快速幂时，需要在ans >= mod时，模完之后还要加上mod;

 

 1 #include <iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 #define ll long long
 5 ll phi(ll c)
 6 {
 7     ll ans = c;
 8     for(int i = 2; i*i <= c; i++) {
 9         if(c%i == 0){
10             ans -= ans/i;
11             while(c%i == 0) c /= i;
12         }
13     }
14     if(c > 1) ans -= ans/c;
15     return ans;
16 }
17 ll quick_mod(ll a, ll b, ll mod)
18 {
19     if(a >= mod) a = a%mod + mod;   // 并不是直接%mod 
20     ll ans = 1;
21     while(b){
22         if(b&1){
23             ans = ans*a;
24             if(ans >= mod) ans = ans%mod + mod;　　//**
25         }
26         a *= a;
27         if(a >= mod) a = a%mod + mod;　　//**
28         b >>= 1;
29     }
30     return ans;
31 }
32 ll solve(ll n, ll m)
33 {
34     ll p = phi(m);
35     if(n == 0) return 1;
36     ll index = solve(n/10, p);
37     
38     return quick_mod(n%10, index, m);
39 }
40 int main()
41 {
42     ll n, m, T;
43     cin >> T;
44     while(T--){
45         scanf("%I64d%I64d", &n, &m);
46         printf("%I64d\n", solve(n,m)%m);
47     }
48     return 0;
49 }

 

posted @ 2016-09-26 20:10  hxer  阅读(725)  评论(0)  编辑  收藏  举报