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.
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).
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
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]