B. Journey Planning
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Tanya wants to go on a journey across the cities of Berland. There are n cities situated along the main railroad line of Berland, and these cities are numbered from 1 to n.
Tanya plans her journey as follows. First of all, she will choose some city c1 to start her journey. She will visit it, and after that go to some other city c2>c1, then to some other city c3>c2, and so on, until she chooses to end her journey in some city ck>ck−1. So, the sequence of visited cities [c1,c2,…,ck] should be strictly increasing.
There are some additional constraints on the sequence of cities Tanya visits. Each city i has a beauty value bi associated with it. If there is only one city in Tanya’s journey, these beauty values imply no additional constraints. But if there are multiple cities in the sequence, then for any pair of adjacent cities ci and ci+1, the condition ci+1−ci=bci+1−bci must hold.
For example, if n=8 and b=[3,4,4,6,6,7,8,9], there are several three possible ways to plan a journey:
c=[1,2,4];
c=[3,5,6,8];
c=[7] (a journey consisting of one city is also valid).
There are some additional ways to plan a journey that are not listed above.
Tanya wants her journey to be as beautiful as possible. The beauty value of the whole journey is the sum of beauty values over all visited cities. Can you help her to choose the optimal plan, that is, to maximize the beauty value of the journey?
Input
The first line contains one integer n (1≤n≤2⋅105) — the number of cities in Berland.
The second line contains n integers b1, b2, …, bn (1≤bi≤4⋅105), where bi is the beauty value of the i-th city.
Output
Print one integer — the maximum beauty of a journey Tanya can choose.
Examples
inputCopy
6
10 7 1 9 10 15
outputCopy
26
inputCopy
1
400000
outputCopy
400000
inputCopy
7
8 9 26 11 12 29 14
outputCopy
55
Note
The optimal journey plan in the first example is c=[2,4,5].
The optimal journey plan in the second example is c=[1].
The optimal journey plan in the third example is c=[3,6].
題意：
題目明顯給出條件，選出的序列必須滿足 Ci+1−Ci=Bci+1−Bci ，求美麗值的最大值。
思路:
- 美麗值i - 美麗值j=下標i - j，很明顯就是美麗值的差值等於下標的差值。
- 通過移項可以得到 Ci+1-Bci+1 = Ci-Bci ，即選出的數Ci對應的會對某個滿足Ci-Bi的序列有所貢獻，所以直接在輸入的時候用map存好就行，最後遍歷map取最大值。
- 注意ci-bi爲負數也可以，昨晚因爲這個點很多人被hack/fst。
代碼:
#include<bits/stdc++.h>
#include<string>
#include<cstring>
#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#define IOS ios::sync_with_stdio(false);cin.tie(0)
#define ll long long
//#define ll unsigned long long
#define inf 0x3f3f3f3f
#define mod 1000000007
#define eps 1e-6
#define PI acos(-1)
#define mysetit multiset<ll>::iterator
#define mymsetit multiset<ll>::iterator
#define mymapit map<ll,ll>::iterator
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define lowbit(x) (x&(-x))
#define mp make_pair
#define pb push_back
#define si size()
#define E exp(1.0)
#define fixed cout.setf(ios::fixed)
#define fixeds(x) setprecision(x)
using namespace std;
inline ll read(){ll s=0,w=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
return s*w;}
void put1(){ puts("Yes") ;}
void put2(){ puts("No") ;}
void put3(){ puts("-1"); }
ll qpf(ll a, ll b, ll p)
{ll ret = 0;while(b){if(b & 1) ret = (ret + a) % p;
a = (a + a) % p;b >>= 1;}return ret % p ;}
ll qp(ll a, ll n, ll p)
{ll ret = 1;while(n){if(n & 1) ret = qpf(ret, a, p);a = qpf(a, a, p);
n >>= 1;}return ret % p ;}
//θ=acos(L/2R);
const int manx=2e5+5;
map<ll,ll>mps;
int main(){
ll n=read();
for(int i=1;i<=n;i++){
ll x=read();
mps[x-i]+=x;
}
ll ans=0;
for(auto &x: mps)
ans=max(ans,x.se);
printf("%lld\n",ans);
return 0;
}