\(\color{#0066ff}{ 题目描述 }\)
给你三个正整数，\(a,m,b\),你需要求： \(a^b \mod m\)
\(\color{#0066ff}{输入格式}\)
一行三个整数，\(a,m,b\)
\(\color{#0066ff}{输出格式}\)
一个整数表示答案
\(\color{#0066ff}{输入样例}\)
2 7 4
998244353 12345 98765472103312450233333333333
\(\color{#0066ff}{输出样例}\)
2
5333
\(\color{#0066ff}{数据范围与提示}\)
注意输入格式，\(a,m,b\) 依次代表的是底数、模数和次数
样例1解释：
\(2^4 \mod 7 = 2\)
输出2
数据范围：
对于全部数据：
\(1≤a≤10^9\)
\(1≤b≤10^{20000000}\)
\(1≤m≤10^6\)
\(\color{#0066ff}{ 题解 }\)
其实就一个式子。。。
\[\left\{\begin{aligned}a^b \equiv a^b \mod m \ \ \ \ \ b < \varphi(m) \\ a^b\equiv a^{b \ mod\ \varphi(m)+\varphi(m)} b\ge\varphi(m)\end{aligned}\right.
\]
#include<bits/stdc++.h>
#define LL long long
LL in() {
char ch; LL x = 0, f = 1;
while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
return x * f;
}
LL ksm(LL x, LL y, LL mod) {
if(x == 0) return 0;
LL re = 1LL;
while(y) {
if(y & 1) re = re * x % mod;
x = x * x % mod;
y >>= 1;
}
return re;
}
int getphi(int x) {
int res = x;
int phi = x;
for(int i = 2; i * i <= x; i++) {
if(res % i == 0) {
while(res % i == 0) res /= i;
phi = phi / i * (i - 1);
}
}
if(res > 1) phi = phi / res * (res - 1);
return phi;
}
int main() {
LL a = in(), m, b;
LL phi = getphi(m = in());
char ch;
while(!isdigit(ch = getchar()));
bool flag = false;
for(b = ch ^ 48; isdigit(ch = getchar()); b = (b << 1LL) + (b << 3LL) + (ch ^ 48)) if(b >= phi) flag = true, b %= phi;
printf("%lld\n", ksm(a, (flag? b + phi : b), m));
return 0;
}
----olinr
