Codeforces Beta Round #17 D. Notepad-优快云博客

本文介绍了一种利用欧拉定理进行模指数运算的方法，并提供了具体的C++实现代码。通过该方法可以有效降低指数运算的时间复杂度，适用于大数运算场景。

主要问题是怎么求$a^b \ mod \ c$。

可以用欧拉定理降幂。

$$a^b \ mod \ c = a ^ {b \% \varphi(c) } \; mod \ c (b < \varphi(c))$$

$$a^b \ mod \ c = a ^ {b \% \varphi(c) + \varphi(c)} \; mod \ c (b \geqslant \varphi(c))$$

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5 char B[1000005], A[1000005];
 6 typedef long long ll;
 7 ll phi(int x) {
 8     ll res = x;
 9     for (int i = 2; i * i <= x; ++i)
10         if (x % i == 0) {
11             res = res / i * (i - 1);
12             while (x % i == 0) x /= i;
13         }
14     if (x > 1) res = res / x * (x - 1);
15     return res;
16 }
17 ll pow(ll a, ll b, ll m) {
18     ll res = 1;
19     for (a %= m; b; b >>= 1, a = (a * a) % m)
20         if (b & 1) res = (res * a) % m;
21     return res;
22 }
23 int c;
24 ll a, a_1, b;
25 int main() {
26     scanf("%s", A); scanf("%s", B); scanf("%d", &c);
27     int lal = strlen(A), lbl = strlen(B);
28     for (int i = 0; i < lal; ++i) {
29         a = (a * 10 + A[i] - '0') % c;
30     }
31     for (int i = lal - 1; ~i; --i) {
32         if (A[i] == '0') A[i] = '9';
33         else {--A[i]; break;}
34     }
35     for (int i = 0; i < lal; ++i)
36         a_1 = (a_1 * 10 + A[i] - '0') % c;
37     int flag = 0, p = phi(c);
38     for (int i = 0; i < lbl; ++i) {
39         b = (b * 10 + B[i] - '0');
40         if (b >= p) flag = 1;
41         b %= p;
42     }
43     if (flag) b += p;
44     ll ans = pow(a, b - 1, c) * a_1 % c;
45     printf("%lld\n", ans ? ans : c);
46     return 0;
47 }