n+1 easy and align acc gre const over color

1003.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 27 思路：貪心，用一個數組記下從i到n的a[i]-i的最大值pre[i],可以得到pre[i]=max(pre[i+1],a[i]-i)，然後容易得到b[k]從小往大取時，後n項之和最大，所以a[n+1]=pre[b[1]]（後n項會呈單調不遞增趨勢，a[n+1]-n-1即為之後取最大值的比較對象）。 代碼：

 1 #include "cstdio"
 2 #include "stdlib.h"
 3 #include "iostream"
 4 #include "algorithm"
 5 #include "string"
 6 #include "cstring"
 7 #include "queue"
 8 #include "cmath"
 9 #include "vector"
10 #include "map"
11 #include "set"
12 #define db double
13 #define ll long long
14 #define inf 0x3f3f3f
15 using namespace std;
16 const int N=3e5+5;
17 const int mod=1e9+7;
18 #define rep(i,x,y) for(int i=x;i<=y;i++)
19 //char s[N],t[N];
20 db  pi=3.14;
21 //int s[N],w[N];
22 int a[N],b[N];
23 int pre[N];
24 int main()
25 {
26     int n;
27     while(scanf("%d",&n)==1){
28         for(int i=1;i<=n;i++){
29             scanf("%d",a+i);
30             a[i]-=i;
31         }
32         memset(pre,0, sizeof(pre));
33         for(int i=n;i>=1;i--) pre[i]=max(a[i],pre[i+1]);
34         for(int i=1;i<=n;i++) scanf("%d",&b[i]);
35         sort(b+1,b+n+1);
36         int ma=0;
37         ll ans=pre[b[1]];
38         for(int i=2;i<=n;i++){
39             ma=max(pre[b[1]]-n-1,pre[b[i]]);
40             ans=(ma+ans)%mod;
41 //            printf("%d\n",ma);
42         }
43         printf("%lld\n",ans);
44     }
45 
46 }

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