Arithmetic Sequence

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+d1and 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

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const long long  maxn = 100005;
long long  A[maxn];
int main()
{
    long long flag,n, d1, d2;
    long long sum, cnt;
    while(~scanf("%I64d %I64d %I64d", &n, &d1, &d2))
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%I64d", &A[i]);
        }

        sum = 0, flag = 0, cnt = 0;

        for(int i = 1; i < n; i++)
        {
            if(A[i]-A[i-1] == d1 && flag == 0)
            {
                cnt ++;
                continue;
            }
            if(A[i]-A[i-1] == d2)
            {
                flag = 1;
                cnt ++;
                continue;
            }
            //printf("cnt = %d\n", cnt);
            sum += (1+cnt)*cnt/2;
            if(cnt != 0) i--;
            cnt = 0;
            flag = 0;
        }
        sum += (1+cnt)*cnt/2;
        //if (d1!=d2)
        printf("%I64d\n", sum + n);
        //else printf("%I64d\n",n);
    }
    return 0;
}

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