OPMODULO - "Operation - Modulo"
Mahmud solved some easy math problems from SPOJ and called himself king of number theory. GodFather GodMATHer Rashad heard it and got angry, so he kidnapped Mahmud. Rashad gave him a task called "Operation - Modulo". Mahmud must solve this task, you know what will happen otherwise ;(. In the Operation - Modulo, we define a function f(n) = (n mod 1) + (n mod 2) + (n mod 3) + ... + (n mod n), where n mod x denotes the remainder when dividing n by x. Rashad interests with integers n such that f(n) = f(n - 1), so he gave Mahmud two numbers L and R, and demands him to find the sum of all integers n such that L ≤ n ≤ R and f(n) = f(n - 1).
Input
First and the only line of input contains two positive integers, L and R (1 ≤ L ≤ R ≤ 1018).
Output
Print the demanded sum in one line.
Example
Input: 1 3 Output: 3
Note: I hope you proved your solution before submitting it :)
hide comments
rajneesh_osho:
2021-08-16 17:56:41
can some one explain it for l=1 and R = 5 |
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aasim3101:
2021-06-18 08:35:26
@mahita_k for the given TC the numbers which are satisfying the condition are 1 and 2 so their sum is 3 which is the answer |
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mahita_k:
2021-06-17 21:33:40
Help me Please
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utkarshag_20:
2021-04-04 19:55:11
nice problem. Last edit: 2021-04-04 19:57:18 |
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sachin_0001:
2021-02-02 10:52:12
I solved it in 0(n^2) ,and got TLE ,please help me in finding optimal approach |
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hodobox:
2019-05-26 03:19:35
@drdrunkenstein, so you see that f(1)=f(0) and f(2)=f(1), so both n=1 and n=2 satisfy f(n)=f(n-1), and the answer is the sum of all such numbers, in this case 1+2 = 3. |
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drdrunkenstein:
2019-04-24 16:58:39
Can someone explain me the test cases?
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sandeep48:
2018-12-25 15:01:15
try to figure out the sequence of no.
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kushagrasri:
2018-11-13 08:04:20
once you get the idea, the problem is super easy.
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abhishak69:
2018-10-06 13:59:40
can you give more testcase
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Added by: | Barish |
Date: | 2018-03-15 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All |
Resource: | Deep places of my brain |