Falling Concrete

Falling Concrete

\(bj = aj - j = aj - (j - i) - i\)
发现不变量
对于一个bi，可以放在任意位置，放在k的值为\(b_i+k\)，要\(b_i+k\)平滑，就要\(b_{i-1} + k_{i-1} = b_i + k_i\)
因为\(k_i = k_{i-1} + 1\), 所以\(b_{i-1} = b_i + 1\)

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

void solve() {
    int n; cin >> n;
    vector<int> b(n);
    for(int i = 0; i < n; i ++) {
        int x; cin >> x;
        b[i] = x - (i + 1);
    }
    sort(b.begin(), b.end());
    b.erase(unique(b.begin(), b.end()), b.end());

    int cur = 1, ans = 1;
    for(int i = 1; i < b.size(); i ++) {
        if(b[i] == b[i-1]+1) cur ++;
        else cur = 1;
        ans = max(cur, ans);
    }
    cout << ans << "\n";
}

int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int _ = 1; cin >> _;
    while(_--) solve();
    return 0;
}