UVA 113 - Power of Cryptography
来源：互联网 发布：淘宝模特入门拍照姿势 编辑：程序博客网 时间：2024/05/18 00:03
题目大意：已知n,p,求k使得k^n=p;
分析：直接pow(p, 1.0/n)，其中p定义为double型
总结：。。。没能理解。。。
代码：
#include <iostream>#include <cstdio>#include <cmath>#define exp 10e-8using namespace std;int main(){int n;double p;while (scanf("%d%lf", &n, &p) != EOF){printf("%d\n",(int)(pow(p, 1.0 / n) + 0.5));}return 0;}- UVa 113 Power of Cryptography
- uva 113Power of Cryptography
- uva 113 Power of Cryptography
- UVA 113 - Power of Cryptography
- uva 113 - Power of Cryptography
- UVa 113 - Power of Cryptography
- UVA 113 - Power of Cryptography
- UVa 113Power of Cryptography
- UVa 113: Power of Cryptography
- uva 113 - Power of Cryptography
- UVA 113 Power of Cryptography
- uva 113 Power of Cryptography
- UVA 113 Power of Cryptography
- uva - 113 - Power of Cryptography
- uva 113 power of Cryptography
- UVa 113 - Power of Cryptography
- uva 113 - Power of Cryptography
- UVa 113 - Power of Cryptography
- 丑女也疯狂
- ubuntu优化
- 绘制透明标签控件(Label)
- 算法大纲
- Android 4.0的图形硬件加速及绘制技巧
- UVA 113 - Power of Cryptography
- 多些时间能少写些代码
- Servlet ---- 基础
- 倒排索引在算法中的运用
- 结构型模式：Facade模式
- 14.2.2 实现和运行颜色滤镜
- HTC G12快速安全root（破解）全攻略(软解s-off root)
- PostgreSQL之精妙的数据库导入导出工具架构 （五）
- VC++.NET 如何更改目标 .Net Framework 的总结
