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(); //} }