hdu 5400 Arithmetic Sequence（模拟）

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

 

 

 

Author

xudyh

 

 

Source

2015 Multi-University Training Contest 9

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 100006
#define ll long long 
int n,k1,k2;
int a[N];
int dp[N];
int main()
{
    while(scanf("%d%d%d",&n,&k1,&k2)==3)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }

        memset(dp,0,sizeof(dp));
        for(int i=1;i<n;i++)
        {
            if(a[i+1]==a[i]+k1)
               dp[i+1]=1;
            else if(a[i+1]==a[i]+k2)
                dp[i+1]=2;
            else dp[i+1]=3;
        }

        ll ans=0;
        ll tmp=0;
        for(int i=1;i<=n;i++)
        {
            if(dp[i]==1)
            {
                if(dp[i-1]==2)
                {
                    tmp=1;
                }
                else
                    tmp++;
                ans=ans+tmp+1;
            }
            else if(dp[i]==2)
            {
                tmp++;
                ans=ans+tmp+1;
            }
            else
            {
                ans++;
                tmp=0;
            }

        }
        printf("%I64d\n",ans);
    }
    return 0;
}