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 = 50005;
int A[N], dp[N];

int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &A[i]);
        }
        int len = 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = n-1; j > i; j--){
                if(j-i < len)break;
                if(A[i] <= A[j]){
                    if(j-i > len)len = j-i;
                }
            }
            if(n-1-i < len)break;
        }
        printf("%d
", len);
    }
    return 0;
}