Codeforces Helpful Maths

最新推荐文章于 2021-09-12 17:43:28 发布

最新推荐文章于 2021-09-12 17:43:28 发布 · 1.1k 阅读

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本文介绍了一个简单的程序设计思路，用于帮助数学初学者Xenia计算加法。通过输入只包含数字1、2、3的算式，程序会重新排列这些数字，使其按非递减顺序排列，从而让Xenia能够轻松地进行计算。

Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.

The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.

You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.

Input

The first line contains a non-empty strings— the sum Xenia needs to count. Stringscontains no spaces. It only contains digits and characters "+". Besides, stringsis a correct sum of numbers 1, 2 and 3. Stringsis at most 100 characters long.

Output

Print the new sum that Xenia can count.

Sample test(s)

input
3+2+1

output
1+2+3

input
1+1+3+1+3

output
1+1+1+3+3

input
2

output
2

这样的题目因为关键字少，所以就可以转换为counting sort的思想去解决。这样时间效率就只有O(n)了。

void HelpfulMaths()
{
	int A[4] = {0};
	int a;
	while (scanf("%d", &a) != EOF)
	{
		A[a]++;
		getchar();
	}
	int total = A[1]+A[2]+A[3]-1;
	for (unsigned i = 0; i < A[1]; i++, total--)
	{
		printf("1");
		if (total) printf("+");
	}
	for (unsigned i = 0; i < A[2]; i++, total--)
	{
		printf("2");
		if (total) printf("+");;
	}
	for (unsigned i = 0; i < A[3]; i++, total--)
	{
		printf("3");
		if (total) printf("+");
	}
}