Codeforces Round #154 (Div. 2) / 253A Boys and Girls(贪心)_there are n boys and m girls studying in the class-优快云博客

本文介绍了CodeForces上的一道经典题目——男孩女孩问题（253A）。该题旨在寻找最优排列方式，使得男孩与女孩交替站立，交替次数尽可能多。文章提供了完整的代码实现，并解释了如何根据不同数量的男孩和女孩来构造最优解。

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There are n boys and m girls studying in the class. They should stand in a line so that boys and girls alternated there as much as possible. Let's assume that positions in the line are indexed from left to right by numbers from 1 to n + m. Then the number of integers i(1 ≤ i < n + m) such that positions with indexes i and i + 1 contain children of different genders (position i has a girl and position i + 1has a boy or vice versa) must be as large as possible.

Help the children and tell them how to form the line.

Input

The single line of the input contains two integers n and m (1 ≤ n, m ≤ 100), separated by a space.

Output

Print a line of n + m characters. Print on the i-th position of the line character "B", if the i-th position of your arrangement should have a boy and "G", if it should have a girl.

Of course, the number of characters "B" should equal n and the number of characters "G" should equal m. If there are multiple optimal solutions, print any of them.

   Sample test(s)
 
     input
   
3 3

     output
   
GBGBGB

     input
   
4 2

     output
   
BGBGBB

water.

完整代码：

/*30ms,100KB*/

#include<cstdio>

int main()
{
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);

	int n, m;
    scanf("%d%d",&n,&m);
	if (n > m)
	{
		int diff = n - m;
		while (m--)
		{
			putchar('B');
			putchar('G');
		}
		while (diff--)
			putchar('B');
	}
	else
	{
		int diff = m - n;
		while (n--)
		{
			putchar('G');
			putchar('B');
		}
		while (diff--)
			putchar('G');
	}
	return 0;
}