POJ2109——Power of Cryptography
2016-08-04 14:37
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题目链接:点击打开链接
题目大意为输入n(1<=n<=200),p(1<=p<10^101)来求一个满足k^n=p公式的k。
由于题目中n,p,k皆为整数,可以考虑用double来存大数,double数据范围有300多位,但是只能精确表示前16位,所以用的时候要保证是整数。转换一下k可以被表示为p^(1/n),这样就可以用pow函数来求k了。
//272K 0MS
//C++ 169B
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
double n,p;
while(cin>>n>>p)
{
cout<<pow(p,1.0/n)<<endl;
}
return 0;
}
题目大意为输入n(1<=n<=200),p(1<=p<10^101)来求一个满足k^n=p公式的k。
由于题目中n,p,k皆为整数,可以考虑用double来存大数,double数据范围有300多位,但是只能精确表示前16位,所以用的时候要保证是整数。转换一下k可以被表示为p^(1/n),这样就可以用pow函数来求k了。
//272K 0MS
//C++ 169B
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
double n,p;
while(cin>>n>>p)
{
cout<<pow(p,1.0/n)<<endl;
}
return 0;
}
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