FZU-1759 Super A^B mod C---欧拉降幂&指数循环节

题目链接：

https://cn.vjudge.net/problem/FZU-1759

题目大意：

求A^B%C

解题思路：

注意，这里long long需要用%I64读入，不能用%lld

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 using namespace std;
 6 typedef long long ll;
 7 const int maxn = 2e6 + 10;
 8 int euler_phi(int n)//求单个
 9 {
10     int m = (int)sqrt(n + 0.5);
11     int ans = n;
12     for(int i = 2; i <= m; i++)if(n % i == 0)
13     {
14         ans = ans / i * (i - 1);
15         while(n % i == 0)n /= i;
16     }
17     if(n > 1)ans = ans / n * (n - 1);
18     return ans;
19 }
20 ll mul(ll a, ll b, ll m)
21 //求a*b%m
22 {
23     ll ans = 0;
24     a %= m;
25     while(b)
26     {
27         if(b & 1)ans = (ans + a) % m;
28         b /= 2;
29         a = (a + a) % m;
30     }
31     return ans;
32 }
33 ll pow(ll a, ll b, ll m)
34 {
35     ll ans = 1;
36     a %= m;
37     while(b)
38     {
39         if(b & 1)ans = mul(a, ans, m);
40         b /= 2;
41         a = mul(a, a, m);
42     }
43     ans %= m;
44     return ans;
45 }
46 ll super_pow(ll a, char b[], ll c)
47 {
48     ll t = euler_phi(c), n = strlen(b), tot = 0;
49     for(int i = 0; i < n; i++)
50     {
51         tot = tot * 10 + b[i] - ‘0‘;
52         tot %= t;
53     }
54     tot += t;
55     return pow(a, tot, c);
56 }
57 char b[maxn];
58 int main()
59 {
60     ll a, c;
61     while(~scanf("%I64d%s%I64d", &a, b, &c))
62     {
63         printf("%I64d\n", super_pow(a, b, c));
64     }
65     return 0;
66 }

原文地址：https://www.cnblogs.com/fzl194/p/9074509.html