TOJ 2926 Series

Description

An arithmetic series consists of a sequence of terms such that each term minus its immediate predecessor gives the same result. For example, the sequence 3, 7, 11, 15 is the terms of the arithmetic series 3+7+11+15; each term minus its predecessor equals 4. (Of course there is no requirement on the first term since it has no predecessor.)

Given a collection of integers, we want to find the longest arithmetic series that can be formed by choosing a sub-collection (possibly the entire collection).

Input

There are multiple cases, and each case contains 2 lines: the first line contains the count of integers (between 2 and 1000 inclusive), the following line contains all the integers (between -1,000,000,000 and 1,000,000,000 inclusive) separated by one or more spaces.

Output

Print a single number for each case in a single line.

Sample Input

7
3 8 4 5 6 2 2
4
-1 -5 1 3
4
-10 -20 -10 -10

Sample Output

5
3
3

Source

ZOJ Monthly, 2005.9

 

看了别人的解题报告说是DP+二分。试着写了一个后来发现与实际的结果相差2，不管三七二十一直接在原来的结果上加2。

竟然过了~o(╯□╰)o

cjx就在想这是为什么呢？后来发现因为我是从第三个数开始处理的，每个运算的结果都会少个2。

 

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <algorithm>
 5 #define MAXN 1001
 6 using namespace std;
 7 
 8 int n,cnt;
 9 int d[MAXN];
10 int a[MAXN];
11 int dp[MAXN][MAXN];
12 
13 int find(int l, int r, int v){    
14     while(l<=r){
15         int m=(l+r)/2;
16         if(v==a[m])return m;
17         else if(v<a[m]){
18             r=m-1;    
19         }else if(v>a[m]){
20             l=m+1;
21         }
22     }
23     return -1;
24 }
25 
26 int main()
27 {
28     int ans,count;
29     int temp;
30      while( scanf("%d",&n)!=EOF ){
31         for(int i=0; i<n; i++){
32             scanf("%d",&d[i]);
33         }
34         sort(d,d+n);
35         cnt=0;
36         count=0;
37         ans=0;
38         temp=-1000000001;
39         for(int i=0; i<n; i++){
40             if(d[i]!=temp){
41                 a[cnt++]=d[i];
42                 if(count>ans)
43                     ans=count;
44                 count=1;
45             }else{
46                 count++;
47             }
48         }
49         if(count>ans){
50             ans=count;
51         }
52         memset(dp,0,sizeof(dp));
53         for(int i=0; i<cnt; i++){
54             for(int j=i+1; j<cnt; j++){
55                 int v=a[j]+a[j]-a[i];
56                 int k=-1;
57                 k=find(j+1,cnt-1,v);
58                 if(k!=-1){
59                     dp[j][k]=dp[i][j]+1;
60                     if(dp[j][k]>=ans){
61                         ans=dp[j][k];
62                     }
63                 }
64             }
65         }
66         printf("%d\n",ans+2);
67      }
68     return 0;
69 }