PAT甲级--1058 A+B in Hogwarts分数 20

题目详情 - 1058 A+B in Hogwarts (pintia.cn)

 

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

测试点分析：

测试点2：不能忘记可能产生溢出。

第一次是把和先转化为Knut，结果测试点2不过。

代码如下：

#include<bits/stdc++.h>
using namespace std;

int main()
{
	int x1, y1, z1, x2, y2, z2;
	scanf("%d.%d.%d %d.%d.%d", &x1, &y1, &z1, &x2, &y2, &z2);
	int ans = x1 * 17 * 29 + y1 * 29 + z1 + x2 * 17 * 29 + y2 * 29 + z2;
	cout << ans / (17 * 29) << "." << ans % (17 * 29) / 29 << "." << (ans % (17 * 29)) % 29;

	return 0;
}

也可以直接使用long long存储数据。

代码如下：

#include<bits/stdc++.h>
using namespace std;

int main()
{
	long long int x1, y1, z1, x2, y2, z2;
	scanf("%lld.%lld.%lld %lld.%lld.%lld", &x1, &y1, &z1, &x2, &y2, &z2);
	long long int ans = x1 * 17 * 29 + y1 * 29 + z1 + x2 * 17 * 29 + y2 * 29 + z2;
	cout << ans / (17 * 29) << "." << ans % (17 * 29) / 29 << "." << (ans % (17 * 29)) % 29;

	return 0;
}

 之后由后向前加并进位。

代码如下：

#include<bits/stdc++.h>
using namespace std;

int main()
{
	int x1, y1, z1, x2, y2, z2;
	scanf_s("%d.%d.%d %d.%d.%d", &x1, &y1, &z1, &x2, &y2, &z2);
	int c = (z1 + z2) % 29;
	int b = (y1 + y2 + (z1 + z2) / 29) % 17;
	int a = x1 + x2 + (y1 + y2 + (z1 + z2) / 29) / 17;
	cout << a << "." << b << "." << c;
	return 0;
}