https://codeforces.com/contest/1321/problem/B
The meaning of problems: the right city n, 1-n, each city there is a b [i] values (Beauty), b make itineraries and maximum values for the number, the city selected itinerary conditions subscript C I +. 1 - C I = B I +. 1 - B I.
Liberation: recursive, since negative numbers map may occur, as can also be zero recursive 4E5 The range b 4E5, open array.
//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j) for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define SC scanf
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
using namespace std;
const int N = 1e6+100;
const int maxn = 1e2+9;
map<int , int>m;
void solve(){
int n ;
cin >> n ;
int years = 0;
rep(i , 1 , n){
int x ;
cin >> x ;
m[x-i] += x ;
years = max (years m [xi]);
}
cout << ans << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie (0); cout.tie (0);
//int t ;
//cin >> t ;
//while(t--){
solve();
//}
}