Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2)——Ddp——Bear and Blocks

http://codeforces.com/contest/574/problem/D

/*
容易得到状态方程
dp[i] = min(a[i], dp[i-1]+1,dp[i-2]+2 ....dp[i+1]+1,,,)
那么可以得到两个递推式
dp[i] = min(a[i], 之间的min+1)
dp[i] = min(a[i], dp[i], 之后的min+1)



*/
/************************************************
* Author        :Powatr
* Created Time  :2015-8-31 15:28:14
* File Name     :cfD.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;

int res[MAXN];
int a[MAXN];
int main(){
    int n;
    while(~scanf("%d", &n)){
        for(int i = 1; i <= n; i++){
            scanf("%d", &a[i]);
            res[i] = a[i];
        }
        
    int now = 0;
    int pre = 0;
    for(int i = 1; i <= n; i++){
        now = pre + 1;
        pre = res[i] = min(res[i], now);
    }
    pre = 0;
    for(int i = n; i >= 1; i--){
        now = pre + 1;
        pre = res[i] = min(res[i], now);
    }
    int max1 = -1;
    for(int i = 1; i <= n; i++)
        max1 = max(max1, res[i]);
    printf("%d\n", max1);
    }
    return 0;
}