A1058. A+B in Hogwarts

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

 1 #include<cstdio>
 2 #include<iostream>
 3 using namespace std;
 4 int main(){
 5     long long A[3], B[3] ,c, c1, c2, c3;
 6     scanf("%lld.%lld.%lld %lld.%lld.%lld", &A[2], &A[1], &A[0], &B[2], &B[1], &B[0]);
 7     c = A[2] * 17 * 29 + A[1] * 29 + A[0] + B[2] * 17 * 29 + B[1] * 29 + B[0];
 8     c1 = c / 17 / 29;
 9     c2 = (c / 29) % 17;
10     c3 = c % 29;
11     printf("%lld.%lld.%lld", c1, c2, c3);
12     cin >> c;
13     return 0;
14 }
View Code

 

1、十分简单的进制转换题，两种方法，统一换算成最低位的单位求和，再换算。或者按照加法进位来做。注意最高位的取值达到了10^7，若用第一种方法，可能要使用long long存储。

2、long long 型在输入和输出时为%lld。

 

posted @ 2018-01-19 17:27  ZHUQW  阅读(97)  评论(0)  编辑  收藏  举报