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 <stdio.h>int main(){int GalleonA,SickleA,KnutA;int GalleonB,SickleB,KnutB;int carry;scanf("%d.%d.%d",&GalleonA,&SickleA,&KnutA);scanf("%d.%d.%d",&GalleonB,&SickleB,&KnutB);carry = (KnutA + KnutB)/29;KnutA = (KnutA + KnutB)%29;SickleA = SickleA + SickleB +carry;carry = SickleA/17;SickleA = SickleA%17;GalleonA = GalleonA +GalleonB +carry;printf("%d.%d.%d\n",GalleonA,SickleA,KnutA);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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