湖北民族学院oj 1883 之 Sequence Number

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本文探讨了在一组有序的不同整数中寻找序列对的最大长度问题，并提供了一段C++实现代码作为解答。

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1883: Sequence Number

Time Limit: 1 Sec Memory Limit: 1280 MB
Total Submissions: 20 Accepted: 6
[ Submit] [ Statistic] [ Go Back]

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.

输入描述

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.

输出描述

Output the answer S in one line for each case.

输入样例

5
2 3 8 6 1

输出样例

AC代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=1e5+10;

int a[maxn];

int main()
{
	int n,maxi;
	while(cin>>n)
	{
	    maxi=0;
		for(int i=0;i<n;i++) cin>>a[i];
		for(int i=n-1;i>=0;i--)
        {
            for(int j=0;j+i<n;j++)
            {
                if(j<i && a[j]<=a[i+j])
                {
                    maxi=max(maxi,i);
                    i=-1;j=n;//退出多重循环
                    break;
                }
            }
        }
        cout<<maxi<<endl;
	}
	return 0;
}