Problem G: Sequence Number----暴力

Problem G: Sequence Number

Time Limit: 1 Sec  Memory Limit: 1280 MB
Submit: 803  Solved: 192
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Description

     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

52 3 8 6 1

Sample Output

3

题目链接：http://acm.hzau.edu.cn/problem.php?cid=1029&pid=6

题目的意思就是新定义一个序列，定义i<j，a[i]<+a[j]，问你最大的j-i的长度是多少。

竟然拿了个一血，我的做法不正规，但是过了，过了。。。

暴力，优化第二层for，特判5,4,3,2,1这种情况。

代码：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int a[100000];int main(){    int n;    while(~scanf("%d",&n)){        bool flag=false;        for(int i=0;i<n;i++){            scanf("%d",&a[i]);            if(i!=0&&a[i]>a[i-1]){                flag=true;            }        }        if(!flag){            printf("0\n");            continue;        }        int ans=0;        int k=0;        for(int i=0;i<n;i++){            k=max(k,i+1);            for(int j=k;j<n;j++){                if(a[j]>=a[i]&&j-i>ans){                    ans=j-i;                    k=j;                }            }        }        printf("%d\n",ans);    }    return 0;}