https://blog.youkuaiyun.com/Nocomplaintszh-cnhttps://blog.youkuaiyun.com/https://blog.youkuaiyun.com/Nocomplaints/article/details/137679836https://blog.youkuaiyun.com/Nocomplaints/article/details/137679836NocomplaintsFri, 12 Apr 2024 14:55:41 +0800https://blog.youkuaiyun.com/Nocomplaints/article/details/137671672https://blog.youkuaiyun.com/Nocomplaints/article/details/137671672NocomplaintsFri, 12 Apr 2024 10:53:33 +0800https://blog.youkuaiyun.com/Nocomplaints/article/details/137650261https://blog.youkuaiyun.com/Nocomplaints/article/details/137650261NocomplaintsThu, 11 Apr 2024 19:37:43 +0800>n>>m)//只要有输入进n,m,就进入循环。while(m–)//注意这里只有区间修改操作是在while内的。//第四步 每个新的a[i]都等同于diff[i]的前缀和。//第二步 定义差分数组(说明差分数组和原数组的关系)//利用前缀和将差分数组 还原为 新的a[i]数组。//第五步 将还原的新的a[i]数组 遍历。]]>https://blog.youkuaiyun.com/Nocomplaints/article/details/137635072https://blog.youkuaiyun.com/Nocomplaints/article/details/137635072NocomplaintsThu, 11 Apr 2024 13:03:24 +0800枚举当前层数搜索的数 会用到。//dfs参数为层数 从第二层开始搜索。//每次搜索结束都返回answer。int dfs(int depth)//按深度进行搜索。//默认输入的数为第一层。//带0的数不算做答案（如0136=136）//第一层已知 不用搜索。]]>https://blog.youkuaiyun.com/Nocomplaints/article/details/137611739https://blog.youkuaiyun.com/Nocomplaints/article/details/137611739NocomplaintsWed, 10 Apr 2024 21:26:02 +0800https://blog.youkuaiyun.com/Nocomplaints/article/details/137610511https://blog.youkuaiyun.com/Nocomplaints/article/details/137610511NocomplaintsWed, 10 Apr 2024 20:50:53 +0800https://blog.youkuaiyun.com/Nocomplaints/article/details/137608468https://blog.youkuaiyun.com/Nocomplaints/article/details/137608468NocomplaintsWed, 10 Apr 2024 19:50:55 +0800https://blog.youkuaiyun.com/Nocomplaints/article/details/136975027https://blog.youkuaiyun.com/Nocomplaints/article/details/136975027NocomplaintsSat, 23 Mar 2024 21:06:15 +0800https://blog.youkuaiyun.com/Nocomplaints/article/details/136971819https://blog.youkuaiyun.com/Nocomplaints/article/details/136971819NocomplaintsSat, 23 Mar 2024 18:05:40 +0800n)//什么时候递归结束。]]>https://blog.youkuaiyun.com/Nocomplaints/article/details/136970446https://blog.youkuaiyun.com/Nocomplaints/article/details/136970446NocomplaintsSat, 23 Mar 2024 16:52:56 +0800https://blog.youkuaiyun.com/Nocomplaints/article/details/136967253https://blog.youkuaiyun.com/Nocomplaints/article/details/136967253NocomplaintsSat, 23 Mar 2024 14:41:46 +0800https://blog.youkuaiyun.com/Nocomplaints/article/details/136952204https://blog.youkuaiyun.com/Nocomplaints/article/details/136952204NocomplaintsFri, 22 Mar 2024 22:07:34 +0800插入元素是无法使用push_back的，因为它的value仅是一个int类型的值而并非容器，map使用insert插入(一对)map中count是记录key(key唯一)的值，count为1说明该key数量为1(key存在)/count为0说明该key数量为0(key不存在)for(auto city:citys)//由于容器中都是不重复的城市,遍历容器,再输出单号。mymap.count§)//如果在mymap中,该城市不存在，说明是第一次出现。//遍历输入，输进map。]]>https://blog.youkuaiyun.com/Nocomplaints/article/details/136952041https://blog.youkuaiyun.com/Nocomplaints/article/details/136952041NocomplaintsFri, 22 Mar 2024 21:09:57 +0800https://blog.youkuaiyun.com/Nocomplaints/article/details/136951343https://blog.youkuaiyun.com/Nocomplaints/article/details/136951343NocomplaintsFri, 22 Mar 2024 20:38:14 +0800使得top为最小的数—->（贪心思想）每次都合并最小的—->每次都让两个最小的数相加—->由于要持续比较，pop出当前最小值，所以相加得到的和要放回。//answer持续在记录，每次只加上最小的。++i)//遍历输入 输入进优先队列。while(pq.size()>=2)//每次相加都需要两个最小的数。//要求队列中至少有两个数，每次只要最小值弹出相加（贪心）//弹出后会更新新的top(最小值)//两个果子相加的结果放回。]]>https://blog.youkuaiyun.com/Nocomplaints/article/details/136949676https://blog.youkuaiyun.com/Nocomplaints/article/details/136949676NocomplaintsFri, 22 Mar 2024 20:11:03 +0800由题目给出的最终的符合所有用例的V(20<=V<=25)，得出不同的B，可根据不同的B分析出符合各个用例的V（如下三种V的范围不同）,最后根据公示推导得出符合各个用例的Ｖ与符合所有用例的V的关系式，然后用符号各个用例的V通过关系式来表示符合所有用例的Ｖ，最后可画图理解寻找最终Ｖ的界。###分析题目——>寻找每个用例的界——>求交集得出符合各个用例的界——>上界与下界即V的最大最小值。//(题目1<=B<=A<=1e9)]]>https://blog.youkuaiyun.com/Nocomplaints/article/details/136949559https://blog.youkuaiyun.com/Nocomplaints/article/details/136949559NocomplaintsFri, 22 Mar 2024 18:57:18 +0800https://blog.youkuaiyun.com/Nocomplaints/article/details/136875250https://blog.youkuaiyun.com/Nocomplaints/article/details/136875250NocomplaintsWed, 20 Mar 2024 14:56:51 +0800计算过的斐波那契数列的那一位就不会再重复计算—->相当于每位只做一次计算操作—->时间复杂度O(n)约为n。原因：上述递归构成一颗搜索数—->结点数约为2^ n个—-> 时间复杂度O(n)约为2^n。//如果这一位已经计算过(demo[n]存在)，那么就直接返回，不再重复计算。输入n，求F（n），n <= 100000，结果对1e9+7取模。n > 3时，F(n) = F(n - 1) + F(n- 2)已知F（1）=F（2）=1；]]>https://blog.youkuaiyun.com/Nocomplaints/article/details/136864775https://blog.youkuaiyun.com/Nocomplaints/article/details/136864775NocomplaintsWed, 20 Mar 2024 12:03:09 +0800i=2—->i=4—->i=8—->i=16—->i=(16+(1+6))=23,i=n,count=5;从1出发，到达15: i=1—->i=2—->i=4—->i=8—->i=16,i>n,count=-1；从1 出发，到达8: i=1—->i=2—->i=4—->i=8,i=n,count=3；比如i=16—->i=(16+(1+6))=23,我们需要知道里面的（1+6）；当前该走的步数i>=目标石头对应的步数n时，说明不可达，输出-1；//每次遍历步数自增1。]]>https://blog.youkuaiyun.com/Nocomplaints/article/details/136852185https://blog.youkuaiyun.com/Nocomplaints/article/details/136852185NocomplaintsTue, 19 Mar 2024 20:44:58 +0800DNA={{‘A’,0},{‘C’,1},{‘G’,2},{‘T’,3}};若字符串up中的i位置与字符串down中的j位置匹配，且字符串up的j位置与字符串的i位置匹配，执行down的i，j位置交换，操作数加1.//序列可能存在不满足操作1的情况，或者执行操作1后序列仍然存在不互补的。（需要进行2次操作1）]]>https://blog.youkuaiyun.com/Nocomplaints/article/details/136849758https://blog.youkuaiyun.com/Nocomplaints/article/details/136849758NocomplaintsTue, 19 Mar 2024 18:49:09 +0800 sum=0时，比如：-1 -1 1 1，虽然乘积不为0，但它的总和却重新回到0，没关系，我们只需要让操作数+1，使得某个数变动，那总和结果就不为0了。举例：1 2 3 0 0 0，为满足以上条件，我们尽量将0都+1 —>1 2 3 1 1 1.0)//（0 0 0）不满足条件进不去，（-1 -1 1 1 ）满足条件可进。0)//比如-1 -1 1 1。]]>