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 + 1 has a boy or vice versa) must be as large as possible.
Help the children and tell them how to form the line.
InputThe single line of the input contains two integers n and m (1 ≤ n, m ≤ 100), separated by a space.
OutputPrint 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 NoteIn the first sample another possible answer is BGBGBG.
In the second sample answer BBGBGB is also optimal.
水题， 优先输出多的内一方，之后间隔输出。
附代码：
1 #include<fstream> 2 using namespace std; 3 int main( void) 4 { 5 int b, g, i=0, m; 6 ifstream fin("input.txt"); 7 ofstream fout("output.txt"); 8 fin>>b>>g; 9 m=b+g; 10 if( b>g) 11 { 12 while( i<m) 13 { 14 if( b>0) 15 { 16 fout<<'B'; 17 b--; 18 i++; 19 } 20 if( g>0) 21 { 22 fout<<'G'; 23 g--; 24 i++; 25 } 26 } 27 } 28 else 29 { 30 while( i<m) 31 { 32 if( g>0) 33 { 34 fout<<'G'; 35 g--; 36 i++; 37 } 38 if( b>0) 39 { 40 fout<<'B'; 41 b--; 42 i++; 43 } 44 } 45 } 46 fout<<endl; 47 return 0; 48 }
转载于:https://www.cnblogs.com/shadow-justice/archive/2013/01/19/2868081.html
