[第一次训练]Arithmetic Progression

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#第一次训练

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本文解决了一个简单的算法问题，即在一个给定序列中找到最长的连续等差子序列。通过逐个比较相邻元素的差值，确定序列是否保持等差性质，从而计算并输出最长子序列的长度。

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Arithmetic Progression

“In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, … is an arithmetic progression with common difference of 2.”

- Wikipedia

This is a quite simple problem, give you the sequence, you are supposed to find the length of the longest consecutive subsequence which is an arithmetic progression.

Input

There are several test cases. For each case there are two lines. The first line contains an integer number N (1 <= N <= 100000) indicating the number of the sequence. The following line describes the N integer numbers indicating the sequence, each number will fit in a 32bit signed integer.

Output

For each case, please output the answer in one line.

Sample Input

1 4 7 9 11 14

Sample Output

解题报告

此题已经无话可说，当时考虑太复杂了。我的理解为，当为1一个数时就是1,2个数就是2，超过两个数至少是2，然后算出后一个差值是否和前一个差值相等，相等则长度加一，不相等即可记录这个长度，待最后比较完成后，找出最长的那段即可。代码如下：

#include<stdio.h>
long long a[100010],ans[1000];
int main()
{
    long long b,x,y,m,l,c,rec;
    int i;
    while(scanf("%lld",&b)!=EOF)
    {
        for(i=0;i<b;i++)
            scanf("%d",&a[i]);
        if(b==1||b==2)
            printf("%lld\n",b);
        else
        {
            m=2;l=0;c=0;
            for(i=0;i<b-2;i++)
            {
                x=a[i+1]-a[i];
                y=a[i+2]-a[i+1];
                if(x==y)
                {
                    m++;
                    l=x;
                }
                else
                {
                    ans[c]=m;
                    m=2;
                    c++;
                }
                ans[c]=m;
            }
            rec=ans[0];
            for(i=1;i<c;i++)
            {
                if(ans[i]>rec)
                    rec=ans[i];
            }
            printf("%lld\n",rec);
        }
    }
    return 0;
}