PAT甲级真题及训练集(14)--1058. A+B in Hogwarts (20)

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

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提交代码

/**作者：一叶扁舟时间：17:41 2017/6/26思路：就是将所有单位的货币钱转换成一种货币单位最小的值,然后使用加法最后再转注:原本都是用int的变量，但是在pat测试时发现总有一个测试不通过，因此后来改为long long结果通过了，说明了如果用int则有可能存在溢出*/#include <stdlib.h>#include <string.h>#include<iostream>using namespace std;int main(){long long pG, pS, pK;long long aG, aS, aK;long long rG, rS, rK;long long  allPK;long long  allAK;long long  rest;//总共多少钱scanf("%lld.%lld.%lld", &pG, &pS, &pK);scanf("%lld.%lld.%lld", &aG, &aS, &aK);//将钱都转成最小单位allPK = pG * 17 * 29 + pS * 29 + pK;allAK = aG * 17 * 29 + aS * 29 + aK;rest = allPK + allAK;rG = rest / (17 * 29);long long temp = rest % (17 * 29);rS = temp / 29;rK = temp % 29;//输出结果printf("%lld.%lld.%lld", rG, rS, rK);system("pause");return 0;}