poj2109-Power of Cryptography
2013-07-22 21:04
435 查看
http://poj.org/problem?id=2109
题意:求a^b = p;
给你b,p求a
#include<stdio.h>
#include<math.h>
int main()
{
double n, m;
while(scanf("%lf%lf", &n, &m) != EOF)
printf("%.lf\n" ,pow(m, 1 / n));
return 0;
}
题意:求a^b = p;
给你b,p求a
#include<stdio.h>
#include<math.h>
int main()
{
double n, m;
while(scanf("%lf%lf", &n, &m) != EOF)
printf("%.lf\n" ,pow(m, 1 / n));
return 0;
}
相关文章推荐
- Power of Cryptography(POJ 2109)(二分)
- POJ 2109 :Power of Cryptography
- poj 2109 Power of Cryptography
- poj 2109 Power of Cryptography
- POJ 2109 Power of Cryptography(水~)
- 二分法+高精度——Poj 2109 Power of Cryptography(double型开n次方的方法通过的原因)
- POJ 2109 Power of Cryptography
- 【二分答案】POJ-2109 Power of Cryptography
- poj 2109 Power of Cryptography
- POJ 2109 - Power of Cryptography
- POJ 2109 Power of Cryptography
- POJ 2109 Power of Cryptography
- poj 2109 Power of Cryptography
- POJ 2109 Power of Cryptography
- poj 2109 Power of Cryptography
- POJ 2109 Power of Cryptography
- poj 2109 : Power of Cryptography
- Poj 2109 / OpenJudge 2109 Power of Cryptography
- poj 2109 Power of Cryptography
- 贪心 POJ 2109 Power of Cryptography