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

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 const long long  maxn = 100005;
 6 long long  A[maxn];
 7 int main()
 8 {
 9     long long flag,n, d1, d2;
10     long long sum, cnt;
11     while(~scanf("%I64d %I64d %I64d", &n, &d1, &d2))
12     {
13         for(int i = 0; i < n; i++)
14         {
15             scanf("%I64d", &A[i]);
16         }
17
18         sum = 0, flag = 0, cnt = 0;
19
20         for(int i = 1; i < n; i++)
21         {
22             if(A[i]-A[i-1] == d1 && flag == 0)
23             {
24                 cnt ++;
25                 continue;
26             }
27             if(A[i]-A[i-1] == d2)
28             {
29                 flag = 1;
30                 cnt ++;
31                 continue;
32             }
33             //printf("cnt = %d\n", cnt);
34             sum += (1+cnt)*cnt/2;
35             if(cnt != 0) i--;
36             cnt = 0;
37             flag = 0;
38         }
39         sum += (1+cnt)*cnt/2;
40         //if (d1!=d2)
41         printf("%I64d\n", sum + n);
42         //else printf("%I64d\n",n);
43     }
44     return 0;
45 }
46
47 [ Copy to Clipboard ]    [ Save to File]

转载于:https://www.cnblogs.com/pblr/p/4739770.html

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