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====== 2020/08/15 -- 2020/08/21 周报 ====== ===== 团队训练 ===== 无 ===== Marvolo ===== ==== 专题 ==== 无 ==== 比赛 ==== [[https://atcoder.jp/contests/abc175|AtCoder Beginner Contest 175]] [[https://codeforces.com/contest/1392|Codeforces Global Round 10 ]] ==== 题目 ==== Codeforces: [[https://codeforces.com/contest/1392/problem/E|Omkar and Duck]] 具体见本周推荐 ===== Kevin ===== ==== 专题 ==== 无 ==== 比赛 ==== 无 ==== 题目 ==== 下次一定 ===== TownYan ===== ==== 专题 ==== 无 ==== 比赛 ==== 无 ==== 题目 ==== 见本周推荐 ===== 本周推荐 ===== ==== Marvolo ==== Codeforces: [[https://codeforces.com/contest/1392/problem/E|Omkar and Duck]] 题意: 交互.先输出一个$n*n$的矩阵,后台会根据这个矩阵,选择一条从左上角到右下角的路线(不可走重复的格子),并输出这条路径上所有数的和.对于后台给出的每个和,要求找到矩阵中对应的路线.要求找出的路线必须和后台找的完全一致. tag: 交互题,构造题 题意: 题目中要求每个格子的数不超过$10^{16}$.已知$2^{50}<10^{16}$.所以,假如说某行的行号是奇数,就把该行全部置为0;否则对于$(i,j)$位置的格子,填入$2^{i+j}$.数据范围保证了$n \leq 25$,刚好卡着题目的上界.画出图后不难发现,路径上的数会依次递增,所以根据后台给的数,按二进制拆位,在矩阵上模拟走一遍即可. comment:很妙的构造题.很想知道tourist是怎么在五分钟内A掉这道题的 ==== Kevin ==== 下次一定 ==== TownYan ==== [[https://ac.nowcoder.com/acm/contest/6915/A]] **题意** 找出[a,b]×[c,d]的点对(x,y)使x×y是质数p的倍数。 **题解** 容斥一下，答案是x是p倍数y任选+y是p倍数x任选-xy都是p倍数