题目来源：2017华中农大网络赛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 eachother. If exist positive integers i , j, ( 1 ≤ i ＜ j ≤ n and A[i] ＞ A[j]), < A[i], A[j]> isregarded as one of A’s inversions. The number of inversions is regarded as inversionnumber. 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 —— equence 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. Thenumber of sequence pairs is regarded as sequence number. Define j – i as the length of thesequence 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） ofthe array.

Output

Output the answer S in one line for each case.

Sample Input

5
2 3 8 6 1

Sample Output

3

题意概括

给出n个数字，问在这n个数字中存在两个数a[i]<=a[j]，求j-i的最大值；

解题思路

首先计算出第一个数字差值，查找第二个数字的时候懂i+max开始查找优化，如果不优化的的话，会超时。

代码

#include <stdio.h>#include <string.h>#include <math.h>#include <ctype.h>#include <queue>#include <algorithm>using namespace std;int a[50010];int main (){    int n,i,j,k,maxx;    while (scanf("%d",&n)!=EOF)    {        k = 0;        for (i = 1; i <= n; i ++)            scanf("%d",a+i);        maxx = 0; j = 1;        for (i = n; i > 0; i --)            if (a[i] >= a[1])                break;        maxx = i-1;        for (j = 2; j <= n-maxx; j ++)        {            for (k = n; k >= j+maxx; k --)            {                if (a[j] <= a[k])                {                    maxx = max(maxx,k-j);                    break;                }            }        }        printf("%d\n",maxx);    }    return 0;}