poj 2109 Power of Cryptography
2011-07-25 20:15
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//此题太犀利了!!!
答案于此: 呵呵呵
#include <iostream> #include<cmath> using namespace std; int main() { double a,b; while (cin>>a>>b) { cout<<pow(b,1.0/a)<<endl; } return 0; }
答案于此: 呵呵呵
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