poj2109-Power of Cryptography
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题意:
对于给出的数开n次方。
思路:
直接用double型数据就可以过,要注意的是pow的第二个参数要写成1.0/n;
#include<iostream>#include<cmath>using namespace std;typedef long long ll;int main(){double n,p;while(cin>>n>>p){printf("%.0lf\n",pow(p,1.0/n));}return 0;}
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