hdu6047

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 856    Accepted Submission(s): 401

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
思路：另建一个数组，来存储前面i 位的最大值，这样就可以用b[i]来扫而且，

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 856    Accepted Submission(s): 401

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
思路：可以再建一个数组来存储前i位的最大值，第n+1位绝对是后面中的最大一个，所以可以直接扫数组b。
代码：
#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <cmath>#include <stdlib.h>#include <vector>#include <queue>#include <stack>using namespace std;const int MOD=1e9+7;vector<int>a1;vector<int>a2;int v[100005];int a[250005],b[250005];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        int i;        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);            a[i]-=i;        }        for(i=1;i<=n;i++)        {            scanf("%d",&b[i]);        }        sort(b+1,b+n+1);        int c[250005];        c[n]=a[n];        for(i=n-1;i>=1;i--)        {            if(a[i]>c[i+1])            {                c[i]=a[i];            }            else c[i]=c[i+1];        }        int d=c[b[1]];        int e=d-n-1;        long long int ans=0;        ans=(ans+d)%MOD;        for(i=2;i<=n;i++)        {            d=c[b[i]];            d=max(e,d);            ans=(ans+d)%MOD;        }        cout<<ans<<endl;    }    return 0;}