int pow_mod(int a, int n, int m){ if(n==0) return 1; int x=pow_mod(a, n/2, m); long long ans = (long long)x*x%m; if(n%2==1)ans=ans*a%m; return (int)ans; } int quick(int a,int b,int c){ int ans = 1; a = a%c; while(b>0){ if(b%2==1)ans=(ans*a)%c;//状态迭代 b= b/2; a = (a*a)%c; } return ans; }
上面的是刘汝佳的写法，复杂度为$O(logn)$