A+B in Hogwarts (20)(模拟)

时间限制 1000 ms 内存限制 65536 KB 代码长度限制 100 KB 判断程序 Standard (来自 小小)

题目描述

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)).

输入描述:

Each input file contains one test case which occupies a line with A and B in the standard form, separated by one space.


 

输出描述:

For each test case you should output the sum of A and B in one line, with the same format as the input.

输入例子:

3.2.1 10.16.27

输出例子:

14.1.28

题解:把进位转换模拟即可,水题

代码:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
	int a1,a2,a3,b1,b2,b3;
	scanf("%d.%d.%d %d.%d.%d",&a1,&a2,&a3,&b1,&b2,&b3);
	int sum1=0,sum2=0,sum3=0;
	
    sum1=a3+b3;
    sum2=b2+a2;
    sum3=a1+b1;
    if(sum1>=29)
    {
    	sum1%=29;
    	sum2++;
	}
	if(sum2>=17)
	{
		sum2%=17;
    	sum3++;
	}
	printf("%d.%d.%d
",sum3,sum2,sum1);
	return 0;
}
原文地址:https://www.cnblogs.com/Staceyacm/p/10782051.html