码迷,mamicode.com
首页 > 其他好文 > 详细

1058. A+B in Hogwarts

时间:2015-02-03 22:39:22      阅读:204      评论:0      收藏:0      [点我收藏+]

标签:

1058. A+B in Hogwarts (20)

时间限制
50 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

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 10.16.27
Sample Output:
14.1.28
 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<stdlib.h>
 4 #include<string.h>
 5 
 6 int main()
 7 {
 8     struct hog
 9     {
10         int g, s, k;
11     };
12     hog a, b, sum;
13     sum.g = sum.s = sum.k = 0;
14     scanf("%d.%d.%d %d.%d.%d", &a.g, &a.s, &a.k, &b.g, &b.s, &b.k);
15     if(a.k + b.k < 29)
16     {
17         sum.k = a.k + b.k;
18     }
19     else
20     {
21         sum.k = a.k + b.k - 29;
22         sum.s++;
23     }
24     if(sum.s + a.s + b.s < 17)
25     {
26         sum.s += a.s + b.s;
27     }
28     else
29     {
30         sum.s += a.s + b.s - 17;
31         sum.g++;
32     }
33     sum.g += a.g + b.g;
34     printf("%d.%d.%d\n", sum.g, sum.s, sum.k);
35     return 0;
36 }

 

1058. A+B in Hogwarts

标签:

原文地址:http://www.cnblogs.com/yomman/p/4271101.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!