1058. A+B in Hogwarts (20)
来源:互联网 发布:舞祭组 知乎 编辑:程序博客网 时间:2024/06/08 15:21
1058. A+B in Hogwarts (20)
If you are a fan of Harry Potter, you would know the world of magic has its own currency system -- as Hagrid explained it to Harry, "Seventeen silver Sickles to a Galleon and twenty-nine Knuts to a Sickle, it's easy enough." Your job is to write a program to compute A+B where A and B are given in the standard form of "Galleon.Sickle.Knut" (Galleon is an integer in [0, 107], Sickle is an integer in [0, 17), and Knut is an integer in [0, 29)).
Input Specification:
Each input file contains one test case which occupies a line with A and B in the standard form, separated by one space.
Output Specification:
For each test case you should output the sum of A and B in one line, with the same format as the input.
Sample Input:3.2.1 10.16.27Sample Output:
14.1.28
#include <iostream>#include <cstdio>using namespace std;int main(){ int h,m,s,h1,m1,s1; scanf("%d.%d.%d",&h,&m,&s); scanf("%d.%d.%d",&h1,&m1,&s1); printf("%d.%d.%d\n",(h+h1+(m+m1+(s+s1)/29)/17),(m+m1+(s+s1)/29)%17,(s+s1)%29);}
- 1058. A+B in Hogwarts (20)- PAT
- 【PAT】1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- PAT 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- PAT 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- pat 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- PAT 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 【2017新疆网络赛】F Islands 强连通分量tarjan hdu2767原题
- 什么是数组
- python super()入门
- 怎么把数据集的输出值转换成只含有0,1的标签向量
- ImageLoad自定义设置
- 1058. A+B in Hogwarts (20)
- EXE made in 2017.9.15.
- 有赞直接打开优惠券中心提示404的解决办法
- SSM框架中的术语
- 正规方程推导过程
- LINTCODE---排序列表转换为二分查找树
- hdu 6194 string string string 后缀数组+rmq+容斥
- SkipList跳跃表
- ARM汇编指令