【题解】Long Jumps 【水题】

链接：https://codeforces.com/contest/1472/problem/C

C. Long Jumps

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp found under the Christmas tree an array a of n elements and instructions for playing with it:

• At first, choose index i(1≤i≤n) — starting position in the array. Put the chip at the index i (on the value ai).
• While i≤n, add aiai to your score and move the chip aiai positions to the right (i.e. replace i with i+ai).
• If i>n, then Polycarp ends the game.

For example, if n=5 and a=[7,3,1,2,3], then the following game options are possible:

• Polycarp chooses i=1. Game process: i=1⟶+78. The score of the game is: a1=7.
• Polycarp chooses i=2. Game process: i=2⟶+35⟶+38. The score of the game is: a2+a5=6.
• Polycarp chooses i=3. Game process: i=3⟶+14⟶+26. The score of the game is: a3+a4=3.
• Polycarp chooses i=4. Game process: i=4⟶+26. The score of the game is: a4=2.
• Polycarp chooses i=5. Game process: i=5⟶+38. The score of the game is: a5=3.

Help Polycarp to find out the maximum score he can get if he chooses the starting index in an optimal way.

Input

The first line contains one integer t(1≤t≤1e4) — the number of test cases. Then tt test cases follow.

The first line of each test case contains one integer n (1≤n≤2e5) — the length of the array aa.

The next line contains nn integers a1,a2,…,an (1≤ai≤1e9) — elements of the array a.

It is guaranteed that the sum of nn over all test cases does not exceed 2e5.

Output

For each test case, output on a separate line one number — the maximum score that Polycarp can get by playing the game on the corresponding array according to the instruction from the statement. Note that Polycarp chooses any starting position from 1 to n in such a way as to maximize his result.

 

题目大意：

一个长度为n的数组a，将一个木块放在1-n任意的起始位置，木块可以向后移动并获得一定的分数。具体规则是：如果一个木块位于位置i，那么获得ai的分数，并移动到i+ai处，如此重复，直到某次i+ai > n，即木块移动出数组。现在已知长度n，和数组中每个元素ai，求所能获得分数的最大值。

题目分析：

假设一个木块在某次移动过程中，到达了位置i，那么它之后移动的路线及之后所能获得的得分只与位置i有关，而与它是如何移动到位置i无关。那么，我们只需要保证木块在到达位置i时已经获得了尽可能多的分数，当木块移动出数组时，获得的分数一定是较优的。

我们只需要维护一个数组b，记录木块移动至i位置时，最多获得bi的分数，那么数组b中的最大值即为我们所求的最后得分的最大值。

代码实现：

#include <bits/stdc++.h>
using namespace std;
const int maxn=2e5+5;
typedef long long ll;
int t;
int n;
int a[maxn];
ll b[maxn];
ll m;
int main(){
    scanf("%d",&t);
    while(t--){
        m=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            b[i]=a[i];
        }
        for(int i=1;i<=n;i++){
            int pos=i+a[i];
            if(pos<=n){
                b[pos]=max(b[pos],b[i]+a[pos]);
            }
            m=max(m,b[i]);
        }
        printf("%lld\n",m);
    }
    return 0;
}