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【偶然一题】CF1979C. Earning on Bets | 思维 | 绕证明绕出来了
bgmaster: 大佬，这个样例跑出来虽然是对的，但是为什么b[i]的和不等于sum 1 2 6 6
关于浏览器打开时自动打开部分网页（浏览器被劫持）的解决办法
zszg: 感谢已解决
Codeforces 1220 D - Alex and Julian
WhiteGrass99: Below is the english translation for the same
Codeforces 1220 D - Alex and Julian
WhiteGrass99: We want to find conditions under which the graph is a bipartite graph, meaning it contains no odd cycles. Therefore, the case of one odd and one even doesn't hold. We aim for both to be odd. To minimize the number of deletions, we express all numbers in B as 2^k * v, where v is an odd number. Only when the exponents k are equal for two numbers, b/gcd(a,b) and a/gcd(a,b) will both be odd. Hence, the retained elements are those with the highest exponents, ranging from 2^0 to 2^63.
Codeforces 1220 D - Alex and Julian
WhiteGrass99: roperty of Bipartite Graph: A graph is bipartite if and only if it does not contain an odd cycle. Utilizing this property, we first consider connecting integers that meet the condition, i.e., when the absolute difference between two integers is in B. For two points a and b existing in B, they can be seen as starting from 0 and advancing forward by a and b steps respectively each time. Therefore, their first meeting point occurs at the least common multiple (LCM) of a and b. At this meeting point, after a has moved b/gcd(a,b) times and b has moved a/gcd(a,b) times, to ensure the absence of an odd cycle, we need the sum of these steps to be divisible by 2. Hence, the problem can be reduced to two cases: Both b/gcd(a,b) and a/gcd(a,b) are odd. One is odd and the other is even. It's impossible for both to be even since if they were both even, it means their denominators can be further divided by 2, contradicting the property of gcd.

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