1058. A+B in Hogwarts (20)
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1058. A+B in Hogwarts (20)
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 10.16.27Sample Output:
14.1.28
#include<iostream>#include<stdio.h>using namespace std;#define LLd long long intint main(){LLd g1, g2;LLd s1, s2;LLd k1, k2;LLd g, s, k;scanf("%lld.%lld.%lld %lld.%lld.%lld", &g1, &s1, &k1, &g2, &s2, &k2);g = g1+g2;s = s1+s2;k = k1+k2;int sDeal = k/29;k = k%29;s = s+sDeal;int gDeal = s/17;s = s%17;g = g+gDeal;printf("%lld.%lld.%lld", g, s, k);return 0;}
- 1058. A+B in Hogwarts (20)- PAT
- 【PAT】1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- PAT 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- PAT 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- pat 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
- PAT 1058. A+B in Hogwarts (20)
- 1058. A+B in Hogwarts (20)
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- 1058. A+B in Hogwarts (20)
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