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Gentleman
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13 years ago,
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How to solve the problem 500 (Div 2) ?
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1a1a1a
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13 years ago,
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Rev. 2
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1 . pre calculate the amount of lucky numbers on segment [0,N]. (you can do this using dp approach). Code will look like this int cnt[4747+1] = {0}; for(int x = 1; x <= 4747; ++x) cnt[x] = cnt[x-1] + isLucky(x); 2. in order to obtain the amount of lucky numbers on arbitrary segment you can use cnt array. number of lucky numbers on segment[M,N] equals to cnt[N] - cnt[N-1]. 3. to solve problem just simulate the game int ans = 0; for(int aa = a; aa + jLen - 1 <= b; ++aa) { int brus = bLen; for(int bb = aa; bb + bLen - 1<= aa + jLen - 1; ++bb) { brus = min(brus, cnt[bb + bLen - 1] - cnt[bb-1] ); } ans = max(ans, brus); } return ans;
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