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CZ_PROB1 - Summing to a Square Prime |
SP2={p∣p:prime∧(∃x1,x2∈Z,p=x21+x22)} is the set of all primes that can be represented as the sum of two squares. The function SP2(n) gives the nth prime number from the set SP2. Now, given two integers n (0<n<501) and k (0<k<4), find p(SP2(n),k) where p(a,b) gives the number of unordered ways to sum to the given total ‘a’ with ‘b’ as its largest possible part. For example: p(5,2)=3 (i.e. 2+2+1, 2+1+1+1, and 1+1+1+1+1). Here 5 is the total with 2 as its largest possible part.
Input
The first line gives the number of test cases T followed by T lines of integer pairs, n and k.
Constraints
- 0<T<501
- 0<n<501
- 1<SP2(n)<7994
- 0<k<4
Output
The p(SP2(n),k) for each n and k. Append a newline character to every test cases’ answer.
Example
Input: 3 2 2 3 2 5 3 Output: 3 7 85
Added by: | Rahul |
Date: | 2007-03-10 |
Time limit: | 1s |
Source limit: | 3000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO NODEJS PERL6 VB.NET |
Resource: | Sam Collins |
hide comments
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2020-04-11 17:37:45
If there is a rendering issue, Read the problem here https://vjudge.net/problem/SPOJ-CZ_PROB1 |
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2019-01-16 23:58:12
There seems to be a problem with the rendering of the mathematical notation. The first line reads: SP2={p∣p:prime∧(∃x1,x2∈Z,p=x21+x22)} is the set of all Could an admin please take a look? |
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2018-05-31 18:37:24
Good question.. |
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2017-06-19 18:39:42
do not forget to consider 2 in your set although its not congruent to 1 modulo 4 |
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2017-06-15 14:05:19
due to small constraints can be solved without dp also ;) #PNC Last edit: 2017-06-15 14:12:20 |
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2017-04-27 14:00:24 shubham
sometimes even the easy ones get you.. Wasted 1.5 hrs in this |
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2017-04-10 04:31:08
Nice Problem :D Fermat's theorem and coin change. |
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2016-11-04 07:16:38
Wrong input constant I fixed 0 < n < 10000 and 1<SP2(n)<10000 ---> AC |
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2016-06-26 14:51:52 surayans tiwari(http://bit.ly/1EPzcpv)
coin change :) |
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2016-06-24 18:47:46
Nyc qsn :) bottom up + precomputation |