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On
prateek_jhunjhunwala →
My rating has stuck around 1620. How do I improve from here on? , 11 years ago
0
Upsolve the problems after the contest and practice those topics which you feel u r weak at. |
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+14
When will the editorial be published? |
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ashish1610 u hacked two div1A problem just 2-3 seconds before I could hack it. |
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I have implemented what has been mentioned in the editorial but it is giving wrong answer on test case #6. Solution Link http://ideone.com/SS4RAe Where am I going wrong? |
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-8
Great tutorial!! PrinceOfPersia can u write a blog on BIT? That would be a lot of help!! |
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0
In 'three sorted arrays' binary search solution : Is the value of t1=[2 1 0] correct? Also while calculating answer 'ans = t1[1]+t1[2]+t1[2] = 3+3+1' why are we doing this? |
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0
Can anyone explain the dynamic programming solution of problem E? |
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+3
Here is the ideone link of the solution http://ideone.com/oGmWhE |
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both one |
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0
Solution to problem 4 was dp + dfs. Here is mine accepted solution of it in codeforces gym. http://codeforces.me/gym/100579/submission/9474632 |
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-45
.. |
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What is the meaning of shortest path in this question? |
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Good tutorial |
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. |
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0
If I want to find distinct increasing subsequence as in this question http://www.spoj.com/problems/INCDSEQ/ what modification I need to make in the above code? |
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0
Merry Christmas to everyone !! Became expert !! |
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0
great explanation !!! thanks.. |
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0
can any anyone explain how is the following relation F(i)=prod(F(j)+1) true? where F(i) be the number of valid sets contain node i as root. F(j) is a node in subtree of i. |
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On
MikeMirzayanov →
2014-2015 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred), 12 years ago
0
Saratov is beautiful !! |
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0
It was a great contest!! |
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0
7 2 1 answer-3 |
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. |
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0
Why don't we divide by the factorial instead of calculating inverse and multiply ? int ans = (f[n] * 1LL / f[k]) % mod; ans = (ans * 1LL /f[n — k]) % mod; |
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0
(E) part int modpow(int a, int b){ int ans = 1; while(b){ if(b & 1) ans = (ans * li(a)) % mod; a = (a * li(a)) % mod; b >>= 1; } return ans; } int rev(int v){ return modpow(v, mod — 2); } Why are we using (mod-2) ? thanks in advance |
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