pat 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

解：开数组存值，存进位，很简单的一题，虽然卡在Galleon位上了，当该位的值大于10000000的时候，我以为可能需要取余，提交发现貌似不需要，于是直接不处理就过了。

代码：

#include<cstdlib>
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    int a[2],b[2],c[2];
    int res[3]; //结果
    int add[3]; //进位
    char ch[4]; //'.'
    cin>>a[0]>>ch[0]>>b[0]>>ch[1]>>c[0];
    cin>>a[1]>>ch[2]>>b[1]>>ch[3]>>c[1];

    add[2]=(c[0]+c[1])/29;
    res[2]=(c[0]+c[1])%29;

    add[1]=(b[0]+b[1]+add[2])/17;
    res[1]=(b[0]+b[1]+add[2])%17;

    res[0]=(a[0]+a[1]+add[1]);
     
    cout<<res[0]<<'.'<<res[1]<<'.'<<res[2]<<endl;
}

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