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IBMOA #62

Minimum Operations to Make Alternating Binary String

估计简单可在线判题

考点(AI 标注)字符串

Each operation flips one bit; make all adjacent characters different and find the minimum number of flips. The empty string already satisfies the condition. The source statement body is missing; the operation was recovered from the title and the original example's bit-by-bit flip explanation. In the original example, flipping the 1st and 3rd bits (1-indexed) of 111101 gives 010101, so the answer is 2; the 101010 written in the original example is a typo for the target string.

输入格式

The first line contains the length n, and the second line contains the binary string (an empty line when n=0). Full original bounds: 0≤n≤200000, and the string contains only 0 and 1.

输出格式

Output one integer, the answer.

样例 1

输入:

5
11101

输出:

1

样例 2

输入:

6
111101

输出:

2

样例 3

输入:

0

输出:

0

样例说明

The outputs of the three public examples are, in order: 1; 2; 0. These answers were verified by independent enumeration or direct simulation.