題目鏈接：http://acm.csu.edu.cn/csuoj/problemset/problem?pid=2021

Description

Everybody loves big numbers (if you do not, you might want to stop reading at this point). There are many ways of constructing really big numbers known to humankind, for instance:

• Exponentiation: 422016=42⋅42⋅...⋅422016 times
  422016=42⋅42⋅...⋅42?2016 times.
• Factorials: 2016!=2016?⋅?2015?⋅?...?⋅?2?⋅?1.

In this problem we look at their lesser-known love-child the exponial, which is an operation defined for all positive integers n? as
exponial(n)=n^{(n?−?1)(n?−?2)?21}
For example, e

Since the exponials are really big, they can be a bit unwieldy to work with. Therefore we would like you to write a program which computes exponial(n

Input

There will be several test cases. For the each case, the input consists of two integers n (1?≤?n?≤?10⁹) and m (1?≤?m?≤?10⁹).

Output

Output a single integer, the value of exponial(n) mod m.

Sample Input

2 42
5 123456789
94 265

Sample Output

2
16317634
39

思路：本題是一道經典的指數循環定理簡記e(n)=exponial(n)e(n)=exponial(n)，利用歐拉定理進行降冪即可，不過要註意會爆int。指數循環公式為指數循環公式為A^B = A^(B %  φ(C) +  φ(C)) % C,其中 φ(C)為1~C-1中與C互質的數的個數。

代碼如下：

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 typedef long long ll;
 5 int n, m;
 6 
 7 ll euler(int n) {
 8     ll ans = n;
 9     for(int i = 2; i * i <= n; i++) {
10         if(n % i == 0) {
11             ans = ans / i * (i - 1);
12             while(n % i == 0) n /= i;
13         }
14     }
15     if(n > 1) ans = ans / n * (n - 1);
16     return ans;
17 }
18 
19 ll ModPow(int x, int p, ll mod) {
20     ll rec = 1;
21     while(p > 0) {
22         if(p & 1) rec = (ll)rec * x % mod;
23         x = (ll)x * x % mod;
24         p >>= 1;
25     }
26     return rec;
27 }
28 
29 ll slove(int n, ll m) {
30     if(m == 1) return 0;
31     if(n == 1) return 1 % m;
32     if(n == 2) return 2 % m;
33     if(n == 3) return 9 % m;
34     if(n == 4) return (1 << 18) % m;
35     return (ll)ModPow(n, euler(m), m) * ModPow(n, slove(n - 1, euler(m)), m) % m;
36 }
37 
38 int main() {
39     while(~scanf("%d%d", &n, &m)) {
40         printf("%lld\n",slove(n, m));
41     }
42     return 0;
43 }

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Exponial （歐拉定理+指數循環定理+歐拉函數+快速冪）