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

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

 using namespace std;

 const int N = ;
 int A[N], dp[N];

 int main()
 {
     int n;
     while(scanf("%d", &n) != EOF)
     {
         for(int i = ; i < n; i++)
         {
             scanf("%d", &A[i]);
         }
         int len = ;
         for(int i = ; i < n; i++)
         {
             for(int j = n-; j > i; j--){
                 if(j-i < len)break;
                 if(A[i] <= A[j]){
                     if(j-i > len)len = j-i;
                 }
             }
             if(n--i < len)break;
         }
         printf("%d\n", len);
     }
     return ;
 }