A1058. A+B in Hogwarts (20)

最新推荐文章于 2024-02-21 16:28:06 发布

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最新推荐文章于 2024-02-21 16:28:06 发布

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本文介绍了一种基于哈利波特系列小说中魔法世界货币系统的计算器实现方法。通过三种不同版本的C/C++代码，展示了如何将两种魔法货币（以加隆、西可和纳特为单位）相加并保持其原有格式输出。此计算器采用模块化设计，易于理解和修改。

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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

题目大意

29个Knut换一个 Sickle
17个Sickle换一个Galleon。
三个版本代码：

Version 1

写的好复杂，好长啊。。。太菜了，还是参考version2吧

#include <stdio.h>
using namespace std;
int main(){
	int array1[3],array2[3];
	scanf("%d.%d.%d %d.%d.%d",&array1[0],&array1[1],&array1[2],&array2[0],&array2[1],&array2[2]);
	int ans[3]={0};
	int ch[3] = {1000001,17,29};
	int temp[3]={0};
	for(int i=2;i>=0;i--){
		if((array1[i]+array2[i])>ch[i]){
			temp[i-1]++;
		}
		temp[i]+=(array1[i]+array2[i])%ch[i];
	}
	printf("%d.%d.%d",temp[0],temp[1],temp[2]);
	return 0;	
}

Version 2

#include <stdio.h>
using namespace std;
int main(){
	int array1[3],array2[3];
	scanf("%d.%d.%d %d.%d.%d",&array1[0],&array1[1],&array1[2],&array2[0],&array2[1],&array2[2]);
	int temp[3]={0};
	int carry=0;
	temp[2]=(array1[2]+array2[2])%29;
	carry = (array1[2]+array2[2])/29;
	temp[1] =(array1[1]+array2[1]+carry)%17;
	carry=(array1[1]+array2[1]+carry)/17;
	temp[0]= (array1[0]+array2[0]+carry);
	printf("%d.%d.%d",temp[0],temp[1],temp[2]);
	return 0;	
}

Version 3:柳神，看了人家的代码，感觉自己写的代码就是渣。自愧不如。

#include <iostream>
using namespace std;
int main() {
    long long a, b, c, d, e, f;
    scanf("%lld.%lld.%lld %lld.%lld.%lld", &a, &b, &c, &d, &e, &f);
    long long num = c + b * 29 + a * 29 * 17 + f + e * 29 + d * 29 * 17;
    long long g = num / (17 * 29);
    num = num % (17 * 29);
    printf("%lld.%lld.%lld", g, num / 29, num % 29);
    return 0;
}