You are given the subsets of indices chosen by Ayush. You need to guess the password. To make a query, you can choose a non-empty subset of indices of the array and ask the maximum of all elements of the array
v[i].clear();intsz;//集合的大小scanf("%d", &sz);intu;for(intj =1; j <= sz; ++j) {scanf("%d", &u); v[i].push_back(u); } }//查询整个数组a的最大值intmx =query(n);//查询mx所在的下标intl =1, r = n;intpos =1;while(l < r) {intmid = l + r >>1;if(query...
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You can even directly calculate this expression by bruteforcing x from 1 to n (and using the fact that maximum distance from a node is one of the diameter endpoints), but ok if you want proof of why x = mid(a, b) call that y, works. If you meet up at say a neighbour of y i...
D. Guess The Maximums(简单二分) D. Guess The Maximums(简单二分) 传送门 思路:题意太迷惑了,它的意思是只要找到一个满足条件的数组P PP即可,而不是要求跟预设的数字一模一样,真实迷惑,显然二分找到最大元素下标所在集合S SS. 然后除了P i P_iPi其他都为最大元素,P i P_iPi的求法就是...
[NACOS HTTP-GET] The maximum number of tolerable server reconnection errors has been reached 2019-12-11 16:50 − 错误的意思是:已达到可容忍的服务器重连接错误的最大数目。有两个解决思路:一个将这个值设置的更大;然后是排查自己连接服务哪儿出了问题。先说在哪儿设置这个值:在拉取nacos服务的注解配...