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pat1058. A+B in Hogwarts (20)

时间:2015-09-01 21:18:53      阅读:260      评论:0      收藏:0      [点我收藏+]

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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

提交代码

 

 1 #include<cstdio>
 2 #include<stack>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<stack>
 6 #include<set>
 7 #include<map>
 8 using namespace std;
 9 int dight[3];
10 int main(){
11     //freopen("D:\\INPUT.txt","r",stdin);
12     int a,b,c;
13     scanf("%d.%d.%d",&a,&b,&c);
14     scanf("%d.%d.%d",&dight[0],&dight[1],&dight[2]);
15     dight[0]+=a;
16     dight[1]+=b;
17     dight[2]+=c;
18     dight[1]+=dight[2]/29;
19     dight[2]%=29;
20     dight[0]+=dight[1]/17;
21     dight[1]%=17;
22     printf("%d.%d.%d\n",dight[0],dight[1],dight[2]);
23     return 0;
24 }

 

pat1058. A+B in Hogwarts (20)

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原文地址:http://www.cnblogs.com/Deribs4/p/4776707.html

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