## page was renamed from 程序设计练习12——zju2329——AB Circle
= AB Circle =

http://acm.zju.edu.cn/show_problem.php?pid=2329

Time limit: 1 Seconds 

Memory limit: 32768K

A circle formed by 'a' and 'b', you are required to figure out all such pairs of positions that when cutting the circle in a pair of the positions, we will get two segments that the number of 'a's in one segment is equal to the number of 'b's in the other segment.

{{attachment:abcircle.gif}}

== Input ==
The length of circle: n is between 2 and 1000 inclusive.

All the characters in the input is either 'a' or 'b'.

There is no strings without any 'a' in the input.

There is no strings without any 'b' in the input.

== Output ==
There are no more than 400 test cases. For each test case, print out all the correct cuttings. For each cutting, print out a pair of integers, representing the two positions to cut on.

In output, all the pairs of positions should be sorted.

The legal positions' number is between 0 and n-1 inclusive.

Print a blank line after each test case.

== Sample Input ==
{{{ab
baa
}}}

== Sample Output ==
=== AB Circle #1: ===
{{{0,1
}}}

=== AB Circle #2: ===
{{{0,1
0,2
1,2
}}}

Note for sample 2

original: baa

cut in 0,1: b aa

the number of 'a's in "b" = the number of 'b's in "aa" = 0

cut in 0,2: ba a

the number of 'b's in "ba" = the number of 'a's in "a" = 1

cut in 1,2: a ab

the number of 'a's in "a" = the number of 'b's in "ab" = 1

BE CAREFULL: the total size of output might be very large.


------

{{{#!cplusplus
/*Written by czk*/
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;

int main() {
	string a;
	int caseno = 0;
	while( cin >> a ) {
		size_t total_a = std::count(a.begin(), a.end(), 'a');
		size_t total_b = std::count(a.begin(), a.end(), 'b');
		if(total_a > total_b) swap(total_a, total_b);
		cout <<"AB Circle #"<< ++caseno <<":"<<'\n';
		for(size_t i = 0; i < total_a + total_b -1; i++) {
			if(total_a <= total_b && i < total_b)
				cout << i <<',' << i+total_a << '\n';
			if(total_a != total_b && i  < total_a )
				cout << i <<',' << i+total_b << '\n' ;
		}
		cout << '\n';
	}
	return 0;
}
}}}