PAT-A+B in Hogwarts (20)
来源:互联网 发布:js转html 编辑:程序博客网 时间:2024/06/05 00:28
题意:
进制的转换
解答:
本题需要注意的是split(“.”)点时,需要转译符//
具体java代码:
import java.util.*;public class Main{ public static void main(String[] args){ Scanner in=new Scanner(System.in); String A=in.next(); String B=in.next(); String[] strA=A.split("\\."); String[] strB=B.split("\\."); int[] intA=new int[3]; int[] intB=new int[3]; for(int i=0;i<3;i++){ intA[i]=Integer.parseInt(strA[i]); intB[i]=Integer.parseInt(strB[i]); } int m=intA[2]+intB[2]; int n=intA[1]+intB[1]; int k=intA[0]+intB[0]; if(m>=29){ m=m-29; n++; } if(n>=17){ n=n-17; k++; } System.out.println(k+"."+n+"."+m); }}
0 0
- 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
- 【c++】PAT(advanced level)1058. A+B in Hogwarts (20)
- App工程结构搭建:几种常见Android代码架构分析
- 一个大神的twitter
- MyBatis入门3--基本操作:增删改+基础查询
- 重新开始
- 矩阵乘法
- PAT-A+B in Hogwarts (20)
- SpringMVC的异常管理
- Java Immutable不可变类
- Linux内核同步机制之(四):spin lock[转]
- 运行异常:已经计划系统关机
- MFC多文档框架
- TOEFL(20151129)分析
- Android性能调优利器StrictMode
- 奇异值分解