首先抠出包围了阵形的最小矩形. 将地图拉伸成一条链,即将第一行.第二行.第三行按顺序连接.阵形也可以用同样的方法处理. 那么问题转化为,给定两个 01 串 S 和 T,问每个 S 中长度为 |T| 的子串是否存在 一个点,两个串对应字符都是 1. 将 T 串翻转,那么就变成了卷积的形式,FFT 计算即可. 在 BFS 求出所有可行的位置之后,对于答案的计算,也是卷积的形式,用 FFT 加速即可. 时间复杂度 O(nm log(nm))…

2019-06-16 14:35:52 1089. Duplicate Zeros - Easy 问题描述: 问题求解: 很显然的可以使用O(n), O(n)的解法.主要问题在于如何在O(1)空间复杂度完成求解. 答案就是通过两次遍历,第一次求出复制后的数组的长度,第二次遍历填写数组. public void duplicateZeros(int[] arr) { if (arr.length == 1) return; final int n = arr.length; int newLen…

链接:https://www.nowcoder.com/acm/contest/141/C 来源:牛客网 题目描述 Eddy likes to play cards game since there are always lots of randomness in the game. For most of the cards game, the very first step in the game is shuffling the cards. And, mostly the randomn…

https://www.nowcoder.com/acm/contest/141#question 一眼背包,用四维dp记录在A,B,C,D条件限制下可以获得的最大知识点,但是题目要求输出路径,在输入中包含0这样的样例,原本的递归寻找路径变的不可行,就需要开五维dp记录在i组条件下ABCD的最大知识点,空间复杂度为36 ^ 5,测试可以通过,但本题有更加优秀的解法,就是在原本四维dp的条件下同时用状压记录已经选择的物品,输出的时候只要输出加入状压的物品即可. #include <map> #i…

链接:https://www.nowcoder.com/acm/contest/141/E 来源:牛客网 题目描述 Eddy likes to play with string which is a sequence of characters. One day, Eddy has played with a string S for a long time and wonders how could make it more enjoyable. Eddy comes up with foll…

链接:https://www.nowcoder.com/acm/contest/141/A 来源:牛客网 Eddy was a contestant participating , Eddy failed to solve a physics equation, which pushed him away from a potential medal. Since then on, Eddy found that physics is actually the most important th…

链接:https://www.nowcoder.com/acm/contest/141/E来源:牛客网 Eddy likes to play with string which is a sequence of characters. One day, Eddy has played with a string S for a long time and wonders how could make it more enjoyable. Eddy comes up with following…

链接:https://www.nowcoder.com/acm/contest/141/H来源:牛客网 Eddy has solved lots of problem involving calculating the number of coprime pairs within some range. This problem can be solved with inclusion-exclusion method. Eddy has implemented it lots of times…

题目链接:https://www.nowcoder.com/acm/contest/141/C 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 262144K,其他语言524288K Special Judge, 64bit IO Format: %lld 题目描述 Eddy likes to play cards game since there are always lots of randomness in the game. For most of the cards g…

题目链接:https://www.nowcoder.com/acm/contest/141/A 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 262144K,其他语言524288K Special Judge, 64bit IO Format: %lld 题目描述 Eddy was a contestant participating in ACM ICPC contests. ACM is short for Algorithm, Coding, Math. Since in…