做题打卡第36天

最新推荐文章于 2025-11-17 20:57:47 发布

原创最新推荐文章于 2025-11-17 20:57:47 发布 · 152 阅读

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这篇博客介绍了如何解决一道编程竞赛题目，该题目涉及动态规划策略。给定一个数组，玩家从特定位置开始，每次将得分累加并移动到下一个位置，目标是最大化总得分。博主分享了他们的解决方案，使用动态规划计算每个位置的最优得分，并给出了C++代码实现。

昨天的竞赛竟然是三个小时！！
我以为是两个点
诶
算了
（以上当我没说）
不狡辩了
就是菜qaq
今天继续总结比赛没做上的题
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 ai to your score and move the chip ai 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≤104) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1≤n≤2⋅105) — the length of the array a.

The next line contains n integers a1,a2,…,an (1≤ai≤109) — elements of the array a.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105.
原题链接：https://codeforces.com/problemset/problem/1472/C
其实就是个递推
想麻烦了
诶
下面是答案：

#include<bits/stdc++.h>

using namespace std;
const int N = 2e5 + 10;
typedef long long ll;

ll t, n;
ll a[N];
ll dp[N];

int main() {

    cin >> t;
    while (t--) {

        cin >> n;
        for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
        ll maxn = 0;
        for (int i = n; i > 0; i--) {
            if(i+a[i]>n){
                dp[i]=a[i];
            }else{
                dp[i]=a[i]+dp[a[i]+i];
            }
            maxn=max(dp[i],maxn);
        }
        cout << maxn << endl;
    }


    return 0;
}