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 10.16.27
Sample Output:
14.1.28
分析:进制转化
code:
#include<iostream>#include<cstdio>using namespace std;int main(){ int a,b,c,d,e,f; scanf("%d.%d.%d %d.%d.%d",&a,&b,&c, &d,&e,&f); int s3=(c+f)%29; int s2=b+e+(c+f)/29; int s1=a+d+s2/17; s2%=17; printf("%d.%d.%d\n",s1,s2,s3); 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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