法1

#include<iostream>
#include<algorithm>
#include<cstring>


using namespace std;

const int N = 1010;

int n;
int a[N];
int res = 0;

int main()
{
    cin>>n;
    for(int i = 0; i<n; i++)
    {
        cin>>a[i];
    }

    int j = 0;
    for(int i = 0; i<n; i = j)
    {
        for(j = i+1; a[i]==a[j]; j++);
        res++;
    }
    cout<<res;
    return 0;
}

法2

#include<iostream>
#include<algorithm>
#include<cstring>


using namespace std;

const int N = 1010;

int n;
int a[N];

int main()
{
    cin>>n;
    for(int i = 0; i<n; i++)  cin>>a[i];

    int res = 1;
    //由观察知：如果当前数与前一个数不一样res++

    for(int i = 1; i<n; i++)
        if(a[i] != a[i-1])
            res++;

    cout<<res;
    return 0;
}