PAT A1058

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

#include "stdio.h"
const int Galleon = 17 * 29;
const int Sickle = 29;
int main(){
    int a1, a2, a3;
    int b1, b2, b3;
    scanf("%d.%d.%d %d.%d.%d", &a1, &a2, &a3, &b1, &b2, &b3);
    long long a, b, c;
    a = a1 * Galleon + a2 * Sickle + a3;
    b = b1 * Galleon + b2 * Sickle + b3;
    c = a + b;
    printf("%lld.%lld.%lld", c / Galleon, c % Galleon / Sickle, c % Sickle);
    return 0;
}

此题与PAT B1037有些类似,但是按照那个写法无法通过(18/20)
看了算法笔记的代码,再百度了一下,还是贴一个我觉得简单的AC吧

#include "stdio.h"
const int Galleon = 17;
const int Sickle = 29;
int main(){
    int a1, a2, a3;
    int b1, b2, b3;
    scanf("%d.%d.%d %d.%d.%d", &a1, &a2, &a3, &b1, &b2, &b3);
    //处理Knuts进位
    a3 += b3;
    a2 += a3 / Sickle;
    a3 %= Sickle;
    //处理Sickle进位
    a2 += b2;
    a1 += a2 / Galleon;
    a2 %= Galleon;
    //处理Galleon
    a1 += b1;

    printf("%d.%d.%d", a1, a2, a3);
    return 0;
}