hdu 5400 Arithmetic Sequence（重构数组）

Arithmetic Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1430    Accepted Submission(s): 625

Problem Description

A sequence  b1,b2,⋯,bn  are called  (d1,d2) -arithmetic sequence if and only if there exist  i(1≤i≤n)  such that for every  j(1≤j<i),bj+1=bj+d1  and for every  j(i≤j<n),bj+1=bj+d2 .

Teacher Mai has a sequence  a1,a2,⋯,an . He wants to know how many intervals  [l,r](1≤l≤r≤n)  there are that  al,al+1,⋯,ar  are  (d1,d2) -arithmetic sequence.

 

Input

There are multiple test cases.

For each test case, the first line contains three numbers  n,d1,d2(1≤n≤105,|d1|,|d2|≤1000) , the next line contains  n  integers  a1,a2,⋯,an(|ai|≤109) .

 

Output

For each test case, print the answer.

 

Sample Input

  
  

   
   5 2 -2
0 2 0 -2 0
5 2 3
2 3 3 3 3
  
  

 

Sample Output

  
  

   
   12
5
  
  

题意：求一个区间，使得区间内存在一点i，序列[l,i)是公差为d1的等差数列，序列[i,r]是公差为d2的等差数列，问你一共有几个区间满足

思路：枚举i点即可，但是直接枚举会有O（n²）的复杂度会炸掉，所以要用l和r数组分别记录当前点最大能延伸到前面和后面的距离，最后相乘即可，注意两点

1：当d1=d2的时候，区间会算重，此时只需要特判即可，值等于所有l或者所有r

2：l和r数组必须为long long 10^6*10^6会超出int型....因为这个WA了好多次。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 100010
#define LL long long
int a[N];
LL l[N],r[N];
int main()
{
    int n,d1,d2;
    while(~scanf("%d %d %d",&n,&d1,&d2))
    {
        long long ans=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            l[i]=r[i]=1;
        }
        for(int i=2; i<=n; i++)
            if(a[i]==a[i-1]+d1)
                l[i]=l[i-1]+1;
        for(int i=n-1; i>=1; i--)
            if(a[i+1]==a[i]+d2)
                r[i]=r[i+1]+1;
        if(d1==d2)
        {
            for(int i=1; i<=n; i++)
                ans+=l[i];
        }
        else
        {
            for(int i=1; i<=n; i++)
                ans+=l[i]*r[i];
        }
        printf("%lld\n",ans);
    }
    return 0;
}