HDU6047 Maximum Sequence（思路）

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 42    Accepted Submission(s): 19

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;

因为b的值不会超过n，因此只需要用一个中间变量刷新最大值即可，然后求i~n区间的最大值，比较一下两者，每次取最大即可

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 250005;const int MOD = 1e9 + 7;int a[N],b[N],c[N];int main(){    int n;    while(~scanf("%d",&n))    {        memset(c,0,sizeof(c));        for(int i=1;i<=n;i++)            {                scanf("%d",&a[i]);                a[i]=a[i]-i;            }        for(int i=n;i>=1;i--)         c[i]=max(a[i],c[i+1]);         int tmp=-1,x;         for(int i=1;i<=n;i++)        scanf("%d",&b[i]);        sort(b+1,b+1+n);        int ans=0;        for(int i=1;i<=n;i++)        {            x=max(c[b[i]],tmp);            ans=((ans%MOD)+(x%MOD))%MOD;            tmp=max(tmp,x-n-i);        }        printf("%d\n",ans);    }}