HDU 6047 Maximum Sequence(多校2)

最新推荐文章于 2018-08-14 09:56:32 发布

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最新推荐文章于 2018-08-14 09:56:32 发布

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分类专栏： HDU 贪心大法 STL Multi-University Training

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本文介绍了一个关于序列问题的算法挑战，任务是确定给定序列的后续部分，通过选择特定值来最大化序列和，并使用优先队列来高效地解决这个问题。

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Problem Description

Steph is extremely obsessed with “sequence problems” that are usually seen on magazines: Given the sequence 11, 23, 30, 35, what is the next number? Steph always finds them too easy for such a genius like himself until one day Klay comes up with a problem and ask him about it.

Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}: an+1…a2n. Just like always, there are some restrictions on an+1…a2n: for each number ai, you must choose a number bk from {bi}, and it must satisfy ai≤max{aj-j│bk≤j

Input

The input contains no more than 20 test cases.
For each test case, the first line consists of one integer n. The next line consists of n integers representing {ai}. And the third line consists of n integers representing {bi}.
1≤n≤250000, n≤a_i≤1500000, 1≤b_i≤n.

Output

For each test case, print the answer on one line: max{∑2nn+1ai} modulo 109+7。

Sample Input

4
8 11 8 5
3 1 4 2

Sample Output

题目大意

给出数组 a 的前 n 项，来求 n+1~2n 的和，后 n 项中 ai 求值的原则是在数组 b 中选取一个数 bk ，然后在数组 a 中选取 $aj-j （bk<=j<i）$ 的最大值最大值作为 ai 项的值。

解题思路

贪心的思想，在求后 n 项和时尽量使当前所求的值最大，那么也就是在数组 a 中选取的范围最大，即在数组 b 中选取的值最小，所以这里需要对数组 b 进行排序；在选取的范围内需要得到 aj-j 的最大值，那么我们可以使用优先队列来存储 aj-j 的值（用数组在更新区间最大值的过程中会超时）；然后每次取出队列中的最大值，判断其 index 是否在选取的 bk 之后范围内，若是，则为当前的 ai 值。

代码实现

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
#define maxn 250007
const int mod=1e9+7;
struct node
{
    int value;
    int index;
    friend bool operator< (node n1, node n2)
    {
        return n1.value < n2.value;
    }
}a;
int b[maxn];
int main()
{
    int n,t;
    while(~scanf("%d",&n))
    {
        priority_queue<node>qu;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&t);
            a.value=t-i;
            a.index=i;
            qu.push(a);
        }
        for(int i=1; i<=n; i++)
            scanf("%d",&b[i]);
        sort(b+1,b+n+1);
        int ans=0;
        node now,ne;
        for(int i=1; i<=n; i++)
        {
            now=qu.top();
            while(now.index<b[i])
            {
                qu.pop();
                now=qu.top();
            }
            ans=(ans+now.value)%mod;
            ne.index=n+i,ne.value=now.value-(n+i);
            qu.push(ne);
        }
        printf("%d\n",ans);
    }
    return 0;
}