CodeForces 1208 A. Boys and Girls 模拟序列

A. Boys and Girls

time limit per test 1 second

memory limit per test 256 megabytes 

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

Note

In the first sample another possible answer is BGBGBG.

In the second sample answer BBGBGB is also optimal.

题目大意： n个男孩, m个女孩， 问(i,i+1) 两个小孩性别不一样最对的情况下的排列， 多种任意输出一种即可

解题思路： 数量大的在外，小的在内， 多出来的放到后面即可。

解题代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char str[300];
int n, m;

int main(){
    freopen( "input.txt", "r", stdin);
    freopen( "output.txt", "w", stdout); 
    while( scanf("%d%d",&n,&m) != EOF)
    {
            int i, flag = n > m ? 1 : 0;    
            int num = 2*min(n,m);    
            for( i = 0; i < num; i++)
            {
                if( flag == 1 ) //boy
                    str[i] = (i&1) ? 'G' : 'B';
                else
                    str[i] = (i&1) ? 'B' : 'G';        
            }
            for(i = num; i < num+max(n,m)-min(n,m); i++)
                str[i] = (flag) ? 'B' : 'G';    
            str[i] = '\0';
            puts(str);        
    }
    return 0;
}