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  • 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;
    }
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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10782051.html
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