Arithmetic Progression解题报告

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

--Wikipedia

 

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

题目大意：给一串数字，算出其中最长等差数列的长度

输入输出要求：

Input

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

 

Output

For each case, please output the answer inone line.

输入输出样例：

Sample Input

6

1 4 7 9 11 14

 

Sample Output

3

 

解题思路：以ans记录最终结果，d为公差，将数组元素一一枚举，l随着d的相等而增加，然后处理其他细节问题即可得到正确结果。

代码：

#include

#define maxn 100000+5

int Array[maxn];

 

int main()

{

       intN;

       while(scanf("%d",&N)==1)

       {

              intans=1;

              ints=0;

              intl,d;

              for(int i=0;i

                     scanf("%d",&Array[i]);

              while(s

              {

                     l=s+1;

                     d=Array[l]-Array[l-1];

                     while(l

                            l++;

                     l--;

                     if(l-s+1>ans)

                            ans=l-s+1;

                     s=l;

              }

              printf("%d\n",ans);

       }    

       return0;

}

解题感想：最开始的想法比较笨，思路大致如下：将所有的公差用一个数组存起来，然后找出公差数组中连续相等的最大值，将其加一即为所求。以样例为例：给出的数组为{1,4,7,9,11,14}，那么对应的公差数组为{3,3,2,2,3}，连续相等的最大值为2，所以输出结果为3，按照这个思路交代码一直编译错误，改成上面的思路后还是编译错误。本来是用const定义maxn的，改成#define就没有编译错误，所以按照之前思路编的代码应该也是对的，可惜之前的代码直接被我删掉了，以后会注意把所有的思路代码都保存好