Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0

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

48 11 8 53 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：

     给定a，b两个数组，其大小相同为n。现在需要找出另一个数组a[n+1]~a[2n]，规定这个数组里的元素满足从b数组种选取一个数b(k)，使得a(i)<=max{a(j)-j} (b(k)<=j<i)，求如何确定这个数组可以达到元素和最大。

思路：

      由于赋值时比较的是a[i]-i的大小，所以肯定是大的值赋在前面，使得a[i]-i尽可能大，这样才能保证选出的元素尽可能大，所得出的和自然也最大。这样我们就先把b数组排序，在a[i]中存入从a[i]到a[n]中a[k]-k的最大值，然后把b从小到大遍历，找到第a[b[i]]个位置再选最大的赋值。下面给组样例。

     4

     11 9 12 5

     1 3 4 3

     a数组做a[k]-k处理为 10 7 9 1

     再做选最大值处理为 10 9 9 1

     b数组排序为1 3 3 4

     则a[5]=10，a[5]-5=5；

        a[6]=9，a[6]-6=3；

        a[7]=9，a[7]-7=2；

        a[8]=5，a[8]-8=-4。

        sum=a[5]+a[6]+a[7]+a[8]=10+9+9+5=23

下面贴上代码：

#include<cstdio>#include<algorithm>#define LL long longusing namespace std;const int MAX=250010;int a[MAX],b[MAX];const LL mod=1e9+7;int main(){    int n;    while(~scanf("%d",&n))    {        int x;        for(int i=1; i<=n; ++i)        {            scanf("%d",&x);            a[i]=x-i;    //把a[i]-i存入数组        }        for(int i=n-1; i>0; --i)            a[i]=max(a[i+1],a[i]); //a[i]代表a[i]~a[n]的最大值        for(int i=1; i<=n; ++i)            scanf("%d",&b[i]);        sort(b+1,b+n+1); //b[0]=0,从b[1]开始排序        LL sum=0;        int most=-1,figure;        for(int i=1; i<=n; ++i)        {            figure=max(most,a[b[i]]);            //a[b[i]]代表从a[b[i]]到a[n],a[k]-k的最大值            //figure(a[n+i+1]的值)取从a[b[i]]到a[n+i],a[k]-k的最大值            sum=(sum+figure)%mod;            most=max(figure-n-i,most);            //figure-n-i代表a[n+i]-(n+i)            //most记录a[n+1]~a[n+i]中的a[k]-k的最大值        }        printf("%lld\n",sum);    }    return 0;}