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ZSUM - Just Add It |
For two given integers n and k find (Zn + Zn-1 - 2Zn-2) mod 10000007, where Zn = Sn + Pn and Sn = 1k + 2k + 3k + … + nk and Pn = 11 + 22 + 33 + … + nn.
Input
There are several test cases (≤ 10000). In each case two space separated positive integers n and k are given.
For last test case n and k are given as 0 0, which is not to be processed.
Constraints
1 < n < 200000000 0 < k < 1000000
Output
For each case print the asked value in separate line.
Example
Input: 10 3 9 31 83 17 5 2 0 0 Output: 4835897 2118762 2285275 3694
Added by: | Manohar Singh |
Date: | 2011-09-04 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | Manohar Singh |
hide comments
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2024-03-09 04:22:31
I am solving from 30 minitues and i came to know that 1e7+7 is given as modulo. |
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2023-11-05 12:24:40
make sure you have read the question properly. that it is saying when to terminate the program -> when you encounter 0 0 stop!! Last edit: 2023-11-06 20:19:10 |
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2023-10-21 11:20:53
can't I use ** in python? Please someone explain me, It says TLE |
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2023-06-23 11:08:09
wtf with mod, i waste so much time with 10^9 + 7 just to realize it is 10^7 + 7 ... |
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2023-02-13 12:12:13
Got it saw the hint where he said use pen and paper lol seems like I am rusty as hell |
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2023-02-06 06:41:07
good question Last edit: 2023-02-06 07:22:17 |
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2022-09-15 15:52:47
Is the equation correct after solving?? Zn+Zn-1 -2Zn-2 = n^k + n^n + (n-1)^k + (n-1)^(n-1) - 2(n-2)^k - 2(n-2)^(n-2) |
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2022-08-04 16:26:41
Hence proved never get afraid just by looking at a question... |
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2021-11-23 20:13:54
very easy, just remember mod = 1e7 + 7, and simply try to reduce the equation by +, - and at the end, use binary exponentiation |
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2021-11-15 16:51:02
Use 10^7 + 7 and not 10^9+7 |