PAT-ADVANCED1058——A+B in Hogwarts

版权声明：我的GitHub：https://github.com/617076674。真诚求星！ https://blog.csdn.net/qq_41231926/article/details/83956112

我的PAT-ADVANCED代码仓：https://github.com/617076674/PAT-ADVANCED

原题链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805416519647232

题目描述：

知识点：进制转换、数据越界

思路：转换为Knut相加，再转换为Galleon.Sickle.Knut的格式输出

需要用long long型变量处理数据，否则数据会越界，测试点2无法通过。

时间复杂度和空间复杂度分析对本题的意义不大。

C++代码：

#include<iostream>
#include<vector>

using namespace std;

long long changeToKnut(long long Galleon, long long Sickle, long long Knut);

int main(){
	long long Galleon1, Sickle1, Knut1, Galleon2, Sickle2, Knut2;
	scanf("%lld.%lld.%lld", &Galleon1, &Sickle1, &Knut1);
	long long AKnuts = changeToKnut(Galleon1, Sickle1, Knut1);
	scanf("%lld.%lld.%lld", &Galleon2, &Sickle2, &Knut2);
	long long BKnuts = changeToKnut(Galleon2, Sickle2, Knut2);
	long long totalKnuts = AKnuts + BKnuts;
	vector<long long> result;
	result.push_back(totalKnuts % 29);
	totalKnuts /= 29;
	result.push_back(totalKnuts % 17);
	result.push_back(totalKnuts / 17);
	printf("%lld.%lld.%lld\n", result[2], result[1], result[0]);
	return 0;
}

long long changeToKnut(long long Galleon, long long Sickle, long long Knut){
	return Galleon * 17 * 29 + Sickle * 29 + Knut;
}

C++解题报告：