topic:
analysis:
The problem is put position . Great idea:
now put the number of small, put 2,5. . . Done and put 0,3 .. . . The rest can be put more.
More simply, directly initialized and more
Code:
char c1='b';
char c2='a';
int n1=B;
int n2=A;
if(A>B)
{
c1='a';
c2='b';
n1=A;
n2=B;
}
string ss(A+B,c1);
//添加n2个c2
int all=(ss.length()-1+1)/3;
//0 1 2 | 3 4 5 | 6 7 5 6 7都归结为 5 x+1 /3 *3
for(int i=2;i<(ss.length()-1+1)/3*3;i=i+3)
{
ss[i]=c2;
}
int c=0;
while(1)
{
if(all==n2) break;
ss[c]=c2;
c=c+3;
all++;
}