SUMMING - SUMMING
Find the sum of $x$ smallest distinct numbers of the series $2^i \times 3^j$ ($i, j \ge 0$).
- the first number of the series is $1 = 2^0 \times 3^0$
- the second number of the series is $2 = 2^1 \times 3^0$
- the third number of the series is $3 = 2^0 \times 3^1$
- the fourth number of the series is $4 = 2^2 \times 3^0$
- the fifth number of the series is $6 = 2^1 \times 3^1$
As the sum can be huge print sum modulo $k$.
Input
The input contains 2 numbers $x$ and $k$: $1 \le x \le 10^{14}$, $1 \le k \le 10^8$
Output
The output contains sum of the first $x$ numbers of the series modulo $k$.
Example
Input: 1 1000 Output: 1
Input: 2 1000 Output: 3
Explanation: $3 = 2^0 \times 3^0 + 2^1 \times 3^0 \pmod{1000}$.
Input: 4 1000 Output: 10
Input: 6 2 Output: 0
Input: 16 1000 Output: 300
hide comments
Amit Digga:
2014-11-21 12:44:24
Ok got it... it says smallest numbers.. |
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Amit Digga:
2014-11-18 19:32:22
what is the 6th term??? |
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Amit Digga:
2014-11-18 19:30:35
if 6th is 2^0*3^2 then i/o are wrongly given |
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abhishek nagpal:
2014-09-04 12:34:03
2590%1000 is 590. Right? -_- |
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Raghav Jajodia:
2014-09-04 10:58:28
@abhishek nagpal: we need to calculate MOD k also. |
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abhishek nagpal:
2014-09-04 06:29:26
The answer for x = 16 k 1000, should not be 590?
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Sumeet Agrawal:
2014-08-06 09:33:26
16 1000 case should give output 333???
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Added by: | priyamehtanit |
Date: | 2013-09-06 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |