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Congrats rng_58 =D |
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Here is a short explanation .. First calculate all sums ( sum[i]= number of subarrays with sum 'i' ) This can be done in O(n^2) Now for each query, go through all sums ( for sum=0 to 2*10^7) and calculate the answer. HINT: for each sum 'i' find how many of s[i]'s will lie between L and R . So the complexity is O(n^2 + (10^7)*Q) . This is just an easy solution (and runs in time . For me it ran in 2.39 mins) . Can anyone discuss a better solution ??! |
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Sorry for my ignorance . But isn't it P!=NP ?? |
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I dont think so. I managed to submit my solution for D (Sad part is it got accepted :( ) . Still don't know whether the tests are final !!! |
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Oh ..One more doubt. Shouldn't dp_buffer[1] be initialised to '1' for each vertex. |
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Yep..Now its fine . |
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Shouldn't "dp_buffer[i + j] += f[v][i]*f[v][j]" (in pseudocode of problem 3) be "dp_buffer[i+j] +=f[cur_node][i]*f[v][j]" ? |
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A great Editorial . Always a great pleasure reading POP's Editorials...Great Job... Really a useful one. Now I feel like I missed a Very Good Round.. Looking forward for ur future contests.. |
