请教测评机相关内容，为什么是WA不是RE

KouMoSir @ 2023-12-23 14:03:53

#include<iostream>
#include<deque>
#include<vector>
#include<cmath>
using namespace std;
typedef long long ll;
const ll N = 2e5 + 10, M = -0x7ffffffffffffff;
ll n, l, r, ans, a[N], f[N], pre[N];
deque<vector<ll>> que;
int main() {
    ans = -0x7fffffff;
    cin >> n >> l >> r;
    for (int i = 0; i <= n; i++)cin >> a[i];
    for (int i = 1; i <= n + l; i++)pre[i] = M;
    for (int i = 0; i <= n + l; i++) {
        if (i > n) {
            ans = max(ans, pre[i]);
            continue;
        }
        while (!que.empty() && que.front()[1] < i + l - r)que.pop_front();
        while (!que.empty() && que.back()[0] <= pre[i])que.pop_back();
        vector<ll> tmp{ pre[i],i };
        que.push_back(tmp);
        pre[i + l] = que.front()[0] + a[i + l];

        //cout << i << " ";
        //for (auto j : que)cout << "{" << j[0] << " " << j[1] << "} ";
        //cout << '\n';
    }
    cout << ans << '\n';
}

const ll N=2e5+10；

const ll N=4e5+10；

by immix @ 2023-12-23 14:09:01

因为下标越界是未定义行为，程序可以做出任何行为

by emo_male_god @ 2023-12-23 14:09:58

@KouMoSir 有时候如果你数组开的大小只比使用的大小小一点点，就可能WA或者TLE，差很多就会RE，还要注意，可能一个数组开小了，但是按照地址的计算，用到了其他数组

by HxDlBbCo877 @ 2024-01-17 20:53:03

因为数组是申请了一块连续的内存，越界的时候访问的是未定义的内存块。然后系统会认为这是和数组一样类型的东西，把它解释成了奇奇怪怪的值，就wa了

by LLLgoyour @ 2024-01-29 08:22:51

看完题解用java写了一遍PriorityQueue解法：

import java.io.*;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class Main {
    static final int[] f = new int[200010];
    static final int[] a = new int[200010];

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int l = Integer.parseInt(st.nextToken());
        int r = Integer.parseInt(st.nextToken());
        st = new StringTokenizer(br.readLine());
        for (int i = 1; i <= n + 1; i++) {
            a[i] = Integer.parseInt(st.nextToken());
        }
        int ans = 0;
        PriorityQueue<Integer> q1 = new PriorityQueue<>((A, B) -> (B - A));
        PriorityQueue<Integer> q2 = new PriorityQueue<>((A, B) -> (B - A));
        for (int i = 1; i <= l; i++) {
            q2.add(a[i]);
        }
        for (int i = l; i <= n; i++) {
            q1.add(f[i - l]);
            if (i - r - 1 >= l) {
                q2.add(f[i - r - 1]);
            }
            while (!q2.isEmpty() && q1.peek().equals(q2.peek())) {
                q1.poll();
                q2.poll();
            }
            f[i] = q1.peek() + a[i];
        }
        for (int i = n - r + 1; i <= n; i++) {
            ans = Math.max(ans, f[i]);
        }
        System.out.println(ans);
    }
}

然后90分。。case #1,#2 (subtask)和#5没过，难道是同样的问题？