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华中农业大学第五届程序设计大赛网络同步赛-G

G. Sequence Number

In Linear algebra, we have learned the definition of inversion number: Assuming A is a ordered set with n numbers ( n > 1 ) which are different from each other. If exist positive integers i , j, ( 1 ≤ i < j ≤ n and A[i] > A[j]), <a[i], a[j]=""> is regarded as one of A’s inversions. The number of inversions is regarded as inversion number. Such as, inversions of array <2,3,8,6,1> are <2,1>, <3,1>, <8,1>, <8,6>, <6,1>,and the inversion number is 5.

Similarly, we define a new notion —— sequence number, If exist positive integers i, j, ( 1 ≤ i ≤ j ≤ n and A[i] <= A[j], <a[i], a[j]=""> is regarded as one of A’s sequence pair. The number of sequence pairs is regarded as sequence number. Define j – i as the length of the sequence pair.

Now, we wonder that the largest length S of all sequence pairs for a given array A.

 

 

Input

There are multiply test cases. In each case, the first line is a number N(1<=N<=50000 ), indicates the size of the array, the 2th ~n+1th line are one number per line, indicates the element Ai (1<=Ai<=10^9) of the array.

Output

Output the answer S in one line for each case.

 

 

Sample Input

5 2 3 8 6 1

 

 

Sample Output

3

 

暴力+剪枝

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 
 8 const int N = 50005;
 9 int A[N], dp[N];
10 
11 int main()
12 {
13     int n;
14     while(scanf("%d", &n) != EOF)
15     {
16         for(int i = 0; i < n; i++)
17         {
18             scanf("%d", &A[i]);
19         }
20         int len = 0;
21         for(int i = 0; i < n; i++)
22         {
23             for(int j = n-1; j > i; j--){
24                 if(j-i < len)break;
25                 if(A[i] <= A[j]){
26                     if(j-i > len)len = j-i;
27                 }
28             }
29             if(n-1-i < len)break;
30         }
31         printf("%d\n", len);
32     }
33     return 0;
34 }

 

posted @ 2017-04-24 12:23  Penn000  阅读(209)  评论(0编辑  收藏  举报