solution

欧拉函数法可以解决模幂运算

#include<stdio.h>
#include<math.h>
int main(){
     int a, m, n, r=1;
     scanf("%d%d%d", &a, &m, &n);
     while(m){
         if(m&1) r=(r*a)%n;
         a=(a*a)%n;
         m>>=1; 
     }
     printf("%d", r);
	return 0;
}

给定伪代码的思想

#include<stdio.h>
#include<math.h>
int isPrime(int n){
	for(int i = 2; i <= sqrt(n); i++){
		if(n % i == 0) return 0;
	}
	return 1;
}
int gcd(int a, int b){
	int temp;
	while(a){
		temp = a;
		a = b % a;
		b = temp;
	}
	return b;
}
int main(){
    int a, m, n, count = 0, c = 1, z[35], num = 0, t;
    scanf("%d%d%d", &a, &m, &n);
    if(isPrime(n) && gcd(a, n) == 1 && m >= n){
    	m%= n-1;
	} 
	t = m;
	do{
		z[num++] = t % 2;
		t /= 2;
		if(z[num-1]) count++; 
	}while(t != 0);
	for(int j = num - 1; j >= 0; j--){
		c = (c*c) % n;
		if(z[j]) c = (a*c) % n;
	}
	printf("%d %d", c, count);
	return 0;
}