n(1e5)个数的数组, 若区间[l, r]上的gcd等于r-l+1, 则该区间就是坏的，每一次操作：选择p，使a[p] = c(c是任何正整数), 对于每个前缀子数组，最少多少次操作使的不存在坏区间

题目

思路：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
const int maxn = 1e6 + 5, inf = 1e9 + 5, maxm = 1e6 + 5;
int a[maxn], b[maxn];
int n, m;
map<int, int> mp;
int suf[maxn], pre[maxn];
bool vis[maxn];
int Gcd[25][maxn], d[maxn];

void init(){
    for(int j = 1; j <= n; j++){
        Gcd[0][j] = a[j];
    }
    for(int i = 1; i <= 20; i++){
        for(int j = 1; j <= n; j++){
            if(j + (1LL << i) - 1 > n) break;
            int t = j + (1LL << (i - 1));
            Gcd[i][j] = __gcd(Gcd[i - 1][j], Gcd[i - 1][t]);
            // cout << i << ' ' << j << ' ' << Gcd[i][j] << '\n';
        }
    }
}
int ask(int l, int r){
    if(r < l) return inf;
    int k = __lg(r - l + 1);
    return __gcd(Gcd[k][l], Gcd[k][r - (1LL << k) + 1]);
}
void solve()
{
    
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    init();
    int j = 1;
    int res = 0;
    int maxr = 0;
    for(int i = 1; i <= n; i++){
        while(j <= i && ask(j, i) < (i - j + 1)){
            j++;
        }
        if(ask(j, i) == i - j + 1){
            if(j > maxr){
                res++;
                maxr = i;
            }
        }  
        cout << res << ' ';
    }
    cout << '\n';
}
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}