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

4 8 11 8 5 3 1 4 2

Sample Output

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;

题目大意：

b序列表示a[ b[i] ]可以用多少次

解题思路：

由样例举例

8 11 8 5

↓ 减去i

7 9 5 1

↓ 因为b数组

I = 1 2 3 4

7 9 5 1

可以使用的次数 1 1 1 1

↓ 因为7<9

I = 1 2 3 4

7 9 5 1

可以使用的次数 0 2 1 1

↓

I = 2 3 4 5

9 5 1 9-i

可以使用的次数 1 1 1 0

ans = ans + 9

↓ 因为9 - i = 4 >1

I = 2 3 4 5

9 5 1 4

可以使用的次数 1 1 0 1

由此下去 ans = 9 + 9 + 5 + 4

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[500004],b[500004];
int n;
long long  modx=1e9+7;
int main()
{while(~scanf("%d",&n)){//memset(a,0,sizeof(a));//memset(b,0,sizeof(b));int k;for(int i=1;i<=n;i++){a[i]=0;b[i]=0;scanf("%d",&a[i]);a[i]=a[i]-i;} for(int i=1;i<=n;i++){scanf("%d",&k);b[k]++; }int yu=a[n],yi=n;for(int i=n-1;i>=1;i--){if(a[i]<yu){        b[yi]+=b[i];b[i]=0;}else{yu=a[i];yi=i;}}//    for(int i=1;i<=n;i++)//{//    printf("%d %d\n",a[i],b[i]);//}int l=n;long long sum=0;int i=1;while(1){if(l==n+n)break;if(b[i]>0){l++;a[l]=a[i]-l;b[l]=1;b[i]--;yu=a[l];yi=l;for(int j=l-1;j>=1;j--){if(a[j]<yu){b[yi]+=b[j];b[j]=0;}elsebreak;}sum=(sum+a[i]+modx)%modx;//printf("%dbvisedhf\n",a[i]);sum%=modx;//printf("%lld\n",sum);}elsei++;}printf("%lld\n",sum);}return 0;
}

2017 杭电多校联赛第二场 1003 Maximum Sequence（单调队列）HDU 6047相关推荐

莫队 ---- 值域分块处理优化字典树上的查询杭电多校2021第二场1004 i love counting
题目链接题目大意: 就是给你nnn个数,mmm次询问每次询问给你l,r,a,bl,r,a,bl,r,a,b问你在[l,r][l,r][l,r]区间里面有多少种种种数c(XOR)a≤bc(XOR)a ...
HDU 5734 Acperience (公式推导) 2016杭电多校联合第二场
题目:传送门. #include <iostream> #include <algorithm> #include <cstdio> #include <cs ...
杭电多校联赛2017年总结
杭电多校联赛2017年总结时间确实过得挺快,一个月很快结束,说实在的,一个月了,每次比赛都是混两道签到题,然后就开始挂机了,确实很不甘心,看着联赛的排名,确实感觉,和强校之前的差距还是很大的,P ...
杭电多校第六场个人补题6 7 9 10 12
杭电多校第六场个人补题6 7 9 10 12 6 题意给定一棵有n结点的树,点权为1~n,求对所有结点子树的mex操作和的最大值思路其实就是从最底部开始网上找,由于0是唯一的一个,所欲最好给在最 ...
2022杭电多校第八场题解
2022杭电多校第八场 Theramore(思维) 题意给定一个01字符串,每次可以将一个奇数长度的区间翻转,求操作后字典序最小的字符串. 分析翻转奇数长度的区间,元素位置的奇偶性不变,统计奇数位 ...
杭电多校第七场 1011 Kejin Player HDU(6656)
杭电多校第七场 1011 Kejin Player 题意:给你N行,代表从i级有花费a[i]元的r[i]/s[i]的概率达到i+1级,剩下的概率中可能会到达x[i]级.然后询问从L级到R级的花费会是多 ...
2019杭电多校第七场 Kejin Player 6656（求期望值）
2019杭电多校第七场 Kejin Player 6656(求期望值) 题目 http://acm.hdu.edu.cn/showproblem.php?pid=6656 题意给你n,q.表示有n ...
2019杭电多校第9场1002 Rikka with Cake HDU6681
2019杭电多校第9场1002 Rikka with Cake HDU6681 题意:给你若干个点按上下左右切割平面,最后问平面内有几块被切割开来. 解法1:红黑树+思维+贪心 A:根据欧拉定理可以得 ...
2019.7.29 杭电多校第三场小结
index > 杭电多校第三场题号标题 AC 做法状态 6603 Azshara's deep sea (51/150)34.00% 6604 Blow up the city (213/ ...

2017 杭电多校联赛第二场 1003 Maximum Sequence（单调队列）HDU 6047

Maximum Sequence

2017 杭电多校联赛第二场 1003 Maximum Sequence（单调队列）HDU 6047相关推荐

最新文章

热门文章