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Codeforces Round 911 (Div. 2)(A-E)思路讲解
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A题:长度大于等于三的区间,可以用两个代价无限水,剩下的情况都要满上。 B题:以a为例子,b和c差距为偶数就是正确,否则错误。 C题:直接dfs,错了就加贡献,最后选择最小。 D题:排序,选择中间数字,后面有多少个大的,就有多少种情况,然后计算a的因子,枚举gcd(a1,a2),最后统一做容斥, E题:如果产生环,那么一定会成为强联通分量,可以缩点,使用tarjan算法,然后得到拓扑序之后进行dp转移,因为只用知道一个缩点的点数量和点权和。
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