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;
}
© 著作权归作者所有