Switch

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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There are N lights in a line. Given the states (on/off) of the lights, your task is to determine at least how many lights should be switched (from on to off, or from off to on), in order to make the lights on and off alternatively.

Input

One line for each testcase.

The integer N (1 <= N <= 10000) comes first and is followed by N integers representing the states of the lights ("1" for on and "0" for off).

Process to the end-of-file.

Output

For each testcase output a line consists of only the least times of switches.

Sample Input

3 1 1 1
3 1 0 1

Sample Output

1
0

这道题很简单，我觉得之所以写是因为提供了自己的思路，在这里将奇数调为开，偶数调为关，

和为num1，这样可以得到另一面就是N-num1就是奇数调为关，偶数调为开，这样哪个结果小于N/2哪个就能成为最小的切换数

#include<stdio.h>

#include<iostream>

#include<cstring>

using namespace std;

int a[10005];

int main()

{

	int i,n,tempo,tempj,temp;

	while(scanf("%d",&n)!=EOF)

	{

		memset(a,0,sizeof(a));

		tempo=tempj=temp=0;

		for(i=0;i<n;i++)

		{

			scanf("%d",&a[i]);

			if(i%2==0)

			{

				if(a[i]==1)

					tempj++;

				else

					tempo++;

			}

			else

			{

				if(a[i]==0)

					tempj++;

				else

					tempo++;

			}

		}

		if(tempj>tempo)

			temp=tempo;

			else

			temp=tempj;

		printf("%d\n",temp);

	}

	return 0;

}