2019独角兽企业重金招聘Python工程师标准>>> 这道题1 <= n <= 200(用int), 1 <= p < 10^101(用double) 调用#include<cmath>函数中的pow(,)函数求解即可 int main() { int n; double p; while(~scanf("%d %lf",&n,&p)) printf("%.0lf\n",pow(p,1.0/n)); return 0; } 转载于:https://my.oschina.net/pandacub/blog/138970