Sequence Number

1570: Sequence Number

时间限制: 1 Sec  内存限制: 1280 MB

题目描述

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.

输入

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.

输出

Output the answer S in one line for each case.

样例输入

5
2 3 8 6 1

样例输出

3

解题思路

求出当A[i]<=A[j],i<=j时，j-i的最大长度。

#include <stdio.h>
#include <algorithm>
using namespace std;
int a[50010];
int main ()
{
    int n, i, j, k, maxn;
    while (~scanf("%d",&n))
    {
        for (i = 0; i < n; i++)
            scanf("%d", &a[i]);
        for (i = n - 1; i >= 0 && a[i] < a[0]; i--);
        maxn = i;
        for (j = 1; j < n - maxn; j++)
        {
            for (k = n - 1; k >= j + maxn; k--)
            {
                if (a[j] <= a[k])
                {
                    maxn = max(maxn, k - j);
                    break;
                }
            }
        }
        printf("%d\n", maxn);
    }
    return 0;
}