Given n , k , find the number of pairs (i,j) such that 0<=i<j<=n and k divides (j-i) Testcases up to 1e5 , (k<=n) up to 1e9 Sample:- input (5,2) , output 6
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3382 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 156 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 145 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 135 |
| 9 | maroonrk | 132 |
| 10 | cry | 130 |
Given n , k , find the number of pairs (i,j) such that 0<=i<j<=n and k divides (j-i) Testcases up to 1e5 , (k<=n) up to 1e9 Sample:- input (5,2) , output 6
| Name |
|---|



Auto comment: topic has been updated by TheSleepyDevil (previous revision, new revision, compare).
i had an idea where the answer is just summation ((n)/k) + ((n-1)/k) + ... till n is k-1 , but idk if that's true or not , or how to calculate it fast
Let j-i=q*k
j-i=x
x=k,2*k,3*k....,(n/k)*k
Let n/k=t (integer division)
n>=j>=x
Number of solutions is n-k+1+(n-2*k+1)+(n-3*k+1)...(n-t+1)=n*t+t-k*t*(t+1)/2
can u explain a bit more ?
UPD: nvm , got it , thanks <3
j-i is multiple of k
All multiples of k<=n are k,2*k,3*k,,,t*k
Let x=q*k
j-i=x
As i>=0 Hence j>=x
also j<=n
Number of such j is n-x+1
x ranges from k,2*k,...t*k