杭电暑期多校集训—Maximum Sequence-优快云博客

本文介绍了一个涉及两个整数序列的问题，目标是找到满足特定条件的序列扩展方式，并求出扩展序列的最大可能和。通过示例解析了算法实现过程，包括输入处理、限制条件和输出结果。

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1887 Accepted Submission(s): 671

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<i}, and any

bk can’t be chosen more than once. Apparently, there are a great many possibilities, so you are required to find max{

∑2nn+1ai} modulo

109+7 .

Now Steph finds it too hard to solve the problem, please help him.

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

27
Hint
For the first sample:
1. Choose 2 from {bi}, then a_2…a_4 are available for a_5, and you can let a_5=a_2-2=9; 
2. Choose 1 from {bi}, then a_1…a_5 are available for a_6, and you can let a_6=a_2-2=9;

题意：根据公式ai≤max{aj-j│bk≤j<i}，对数列{ai},{bi},进行处理
用样例来分析
a： 8 11 8 5
a[i]-i: 7 9 5 1
b：3 1 4 2
先将b排序得到 b: 1 2 3 4
那么取第一个b[0]=1 那么从a1开始找，取最大值9
a：8 11 8 5 9
a[i]-i: 7 9 5 1 4
b[1]=2 从a2开始找，还是取到 9
a: 8 11 8 5 9 9
a[i]-i: 7 9 5 1 4 3
后面继续这样取最终得到的是
a: 8 11 8 5 9 9 5 4 多出来的就是 27

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int a[3000020];
int b[250005];
int c[230005];
int main()
{
    long long int n;
    int ds=0;
    int Mo = 1e9 + 7;
    while(~scanf("%lld",&n))
    {
        long long int ans=0,sum=0;
        ds++;
        if(ds>20)
            break;
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
            scanf("%d",&b[i]);
        sort(b+1,b+n+1);
        for(int i=n;i>0;i--)
        {
            if(a[i]-i>=ans)
                ans=a[i]-i;
            c[i]=ans;
        }
        int max1=c[1]-n-1;
        for(int i=1;i<=n;i++)
        {
            if(c[b[i]]>max1)
                a[n+i]=c[b[i]];
            else
                a[n+i]=max1;
            if(a[n+i]-n-i>max1)
                max1=a[n+i]-n-i;
            sum+=a[n+i];
            if(sum>=Mo)
                sum%=Mo;
        }
        printf("%lld\n",sum);
    }
    return 0;
}