DIV + MOD solution codeforces
Not so long ago, Vlad came up with an interesting function:
- 𝑓𝑎(𝑥)=⌊𝑥𝑎⌋+𝑥mod𝑎fa(x)=⌊xa⌋+xmoda, where ⌊𝑥𝑎⌋⌊xa⌋ is 𝑥𝑎xa, rounded down, 𝑥mod𝑎xmoda — the remainder of the integer division of 𝑥x by 𝑎a.
For example, with 𝑎=3a=3 and 𝑥=11x=11, the value 𝑓3(11)=⌊113⌋+11mod3=3+2=5f3(11)=⌊113⌋+11mod3=3+2=5.
The number 𝑎a is fixed and known to Vlad. Help Vlad find the maximum value of 𝑓𝑎(𝑥)fa(x) if 𝑥x can take any integer value from 𝑙l to 𝑟r inclusive (𝑙≤𝑥≤𝑟l≤x≤r).
DIV + MOD solution codeforces
The first line of input data contains an integer 𝑡t (1≤𝑡≤1041≤t≤104) — the number of input test cases.
This is followed by 𝑡t lines, each of which contains three integers 𝑙𝑖li, 𝑟𝑖ri and 𝑎𝑖ai (1≤𝑙𝑖≤𝑟𝑖≤109,1≤𝑎𝑖≤1091≤li≤ri≤109,1≤ai≤109) — the left and right boundaries of the segment and the fixed value of 𝑎a.
For each test case, output one number on a separate line — the maximum value of the function on a given segment for a given 𝑎a.
5 1 4 3 5 8 4 6 10 6 1 1000000000 1000000000 10 12 8
DIV + MOD solution codeforces
2 4 5 999999999 5
DIV + MOD solution codeforces
In the first sample:
- 𝑓3(1)=⌊13⌋+1mod3=0+1=1f3(1)=⌊13⌋+1mod3=0+1=1,
- 𝑓3(2)=⌊23⌋+2mod3=0+2=2f3(2)=⌊23⌋+2mod3=0+2=2,
- 𝑓3(3)=⌊33⌋+3mod3=1+0=1f3(3)=⌊33⌋+3mod3=1+0=1,
- 𝑓3(4)=⌊43⌋+4mod3=1+1=2f3(4)=⌊43⌋+4mod3=1+1=2
As an answer, obviously, 𝑓3(2)f3(2) and 𝑓3(4)f3(4) are suitable.