Python 实现 PAT:A+B in Hogwarts (20)
来源：互联网 发布：淘宝网签证 编辑：程序博客网 时间：2024/05/18 17:26
A+B in Hogwarts (20)
这题比较简单，P
#PYTHON 3.5AB=input().split(' ')A= [int(x) for x in AB[0].split('.')]B= [int(x) for x in AB[1].split('.')]C= [x+y for x,y in zip(A ,B)]if C[2]>=29: C[2]-=29 C[1]+=1if C[1]>=17: C[1]-=17 C[0]+=1print('%d.%d.%d'%(C[0],C[1],C[2]))阅读全文
0 0
- Python 实现 PAT:A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)- PAT
- 【PAT】1058. A+B in Hogwarts (20)
- PAT 1058. A+B in Hogwarts (20)
- PAT 1058. A+B in Hogwarts (20)
- pat 1058. A+B in Hogwarts (20)
- PAT 1058. A+B in Hogwarts (20)
- PAT 1058. A+B in Hogwarts (20)
- PAT-A+B in Hogwarts (20)
- PAT 1018A+B in Hogwarts (20)
- 【PAT】A1058. A+B in Hogwarts (20)
- PAT---1058. A+B in Hogwarts (20)
- PAT A1058. A+B in Hogwarts (20)
- PAT A 1058. A+B in Hogwarts (20)
- PAT(A) - 1058. A+B in Hogwarts (20)
- PAT-A 1058. A+B in Hogwarts (20)
- PAT-A-1058. A+B in Hogwarts (20)
- PAT-A 1058. A+B in Hogwarts
- TensorFlow Wide & Deep Learning 中遇到的bug 主要是为了跑通Wide & Deep Learning。 环境： VMware ubuntu python 3.5
- JavaScript学习（二）：原型继承
- 线性代数->向量、方程组总结
- s-域与z-域的各种离散方法区别分析
- 嵌入式系统移植
- Python 实现 PAT:A+B in Hogwarts (20)
- HDOJ 2135 Rolling table
- 阶段小结
- 四、使用账号和密码远程连接实例
- 数字序列
- 嵌入式和单片机的区别
- 沉浸式
- 如何创建安装和启用Samba服务
- php编程文件上传代码
