1058. A+B in Hogwarts (20)-PAT甲级真题
2016-08-04 17:32
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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
题目大意:17个Sickle对换一个Galleon,29个Knut对换一个Sickle。根据Galleon.Sickle.Knut的方式相加A和B
分析:把A和B都化为Knut的形式,然后相加,最后转换为Galleon.Sickle.Knut的形式。
注意:A和B相加有可能超出int范围,使用long
long格式存储数据。
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
题目大意:17个Sickle对换一个Galleon,29个Knut对换一个Sickle。根据Galleon.Sickle.Knut的方式相加A和B
分析:把A和B都化为Knut的形式,然后相加,最后转换为Galleon.Sickle.Knut的形式。
注意:A和B相加有可能超出int范围,使用long
long格式存储数据。
#include <cstdio> using namespace std; int main() { long long a, b, c, d, e, f; scanf("%lld.%lld.%lld %lld.%lld.%lld", &a, &b, &c, &d, &e, &f); long long num = c + b * 29 + a * 29 * 17 + f + e * 29 + d * 29 * 17; long long g = num / (17 * 29); num = num % (17 * 29); printf("%lld.%lld.%lld", g, num / 29, num % 29); return 0; }
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