zoj 2486 Power of Cryptography
2007-02-26 11:01
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#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int main()
...{
double n,p;
while(scanf("%lf%lf",&n,&p)!=EOF)
...{
double k = pow(p, 1.0/n);
printf("%g ",k);
}
return 0;
}
这道题很简单,用pow() function
用库函数,要注意的是把开方次数转换成实型
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