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 <algorithm>#include <stdio.h>#include <map>using namespace std;int main() { string str1,str2; int a1,a2,a3,b1,b2,b3,a,b,c; cin>>str1>>str2; sscanf(str1.c_str(),"%d.%d.%d",&a1,&a2,&a3); sscanf(str2.c_str(),"%d.%d.%d",&b1,&b2,&b3); a=a1+b1; b=a2+b2; c=a3+b3; b=b+c/29; c=c%29; a=a+b/17; b=b%17; printf("%d.%d.%d\n",a,b,c); 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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