多校二 1003Maximum Sequence 模拟

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：在a数组后面添加n个数， 其中每个数需要满足它是从bk的位置到a数组最后中小于等于

aj-j的值，求所添加的数的和的最大值。

 

分析：样例分析：

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 那么取到9

a：8 11 8 5 9

a[i]-i: 7 9 5 1 4

b[1]=2 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

 

那么从模拟的这个过程可以发现，只要记录当前位置到n中间的最大值 和更新出来的数中的最大值

进行比较取最大的那个，就是答案啦~

 

 

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<iostream>
#include<algorithm>

using namespace std;
#define N 300009
#define mod 1000000007
long long a[N], b[N], maxa[N];
int main()
{

int n;

while(scanf("%d", & n) != EOF)
{
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
memset(maxa, 0, sizeof(maxa));

for(int i = 1; i <= n; i++)
{
scanf("%lld", &a[i]);
a[i] = a[i] - i;
}

for(int i = 1; i <= n; i++)
scanf("%lld", &b[i]);

sort(b, b+n+1);

for(int i = n ; i >= 1; i--)
{
maxa[i] = max(maxa[i+1], a[i]);
}

long long ans = 0;

ans += maxa[b[1]];
ans %= mod;

long long t = maxa[b[1]] - n - 1;

for(int i = 2; i <= n; i++)
{
maxa[b[i]] = max(maxa[b[i]], t);

ans += maxa[b[i]];
ans %= mod;

t = max(t, maxa[b[i]] - n - i);

}
printf("%lld\n", ans);
}
}