题目分析
题目的本意就是告诉你一串序列，根据这串序列去判断另一个序列的有多少种不同的值

方法一：模拟

直接按照题目的意思模拟即可，但是要注意先后顺序，也就是说循环遍历的顺序，最后我们直接用set统计一下有多少个不同的值就是答案

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <set>
#include <map>
#include <math.h>
#include <vector>
#include <queue>
#include <string.h>
typedef long long ll;
using namespace std;
#define pi acos(-1.0)
const int maxn = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int a[maxn], c[maxn];
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d", &c[i]);
        a[i] = i;
    }
    for (int i = 2; i <= n; ++i)
        if (c[i] == 0)
            a[i] = a[i - 1];
    for (int i = n - 1; i >= 1; --i)
        if (c[i])
            a[i] = a[i + 1];
    set<int> s;
    for (int i = 1; i <= n; ++i)
        s.insert(a[i]);
    printf("%d\n", s.size());
}

方法二：找规律

#include <bits/stdc++.h>
using namespace std;

#define maxn 500005

int n, m, ans = 1;
int a[maxn];

int main()
{

    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i < n; i++)
        if (a[i] == 0 && a[i + 1] == 1)
            ans++;
    cout << ans;
    return 0;
}