<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[qq_44527245的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/qq_44527245</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; qq_44527245]]></copyright><item><title><![CDATA[Enterprising Escape]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/116451252</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/116451252</guid><author>qq_44527245</author><pubDate>Thu, 06 May 2021 12:33:42 +0800</pubDate><description><![CDATA[Enterprising Escape

The Enterprise is surrounded by Klingons! Find the escape route that has the quickest exit
time, and print that time.
Input is a rectangular grid; each grid square either has the Enterprise or some class of a Klingon
warship. Ass...]]></description><category></category></item><item><title><![CDATA[Expect the Expected]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/111622730</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/111622730</guid><author>qq_44527245</author><pubDate>Thu, 24 Dec 2020 13:29:51 +0800</pubDate><description><![CDATA[Some mathematical background. This problem asks you to compute the expected value of a random
variable. If you haven’t seen those before, the simple definitions are as follows. A random variable is a
variable that can have one of several values, each with ]]></description><category></category></item><item><title><![CDATA[Tribles]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/111604972</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/111604972</guid><author>qq_44527245</author><pubDate>Thu, 24 Dec 2020 10:04:26 +0800</pubDate><description><![CDATA[GRAVITATION, n.
“The tendency of all bodies to approach one another with a strength
proportion to the quantity of matter they contain – the quantity of
matter they contain being ascertained by the strength of their tendency
to approach one another. This is]]></description><category></category></item><item><title><![CDATA[白兔的字符串]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/111600711</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/111600711</guid><author>qq_44527245</author><pubDate>Wed, 23 Dec 2020 22:30:22 +0800</pubDate><description><![CDATA[链接：https://ac.nowcoder.com/acm/problem/15253
来源：牛客网
白兔有一个字符串T。白云有若干个字符串S1,S2…Sn。
白兔想知道，对于白云的每一个字符串，它有多少个子串是和T循环同构的。
提示：对于一个字符串a，每次把a的第一个字符移动到最后一个，如果操作若干次后能够得到字符串b，则a和b循环同构。
所有字符都是小写英文字母
第一行一个字符串T（|T|&lt;=10^6）
第二行一个正整数n (n&lt;=1000)
接下来n行为S1~Sn (|S1|+|S2|+]]></description><category></category></item><item><title><![CDATA[Investigating Div-Sum Property]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/111403204</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/111403204</guid><author>qq_44527245</author><pubDate>Sat, 19 Dec 2020 13:28:39 +0800</pubDate><description><![CDATA[An integer is divisible by 3 if the sum of its digits is also divisible by 3. For example, 3702 is divisible
by 3 and 12(3+7+0+2) is also divisible by 3. This property also holds for the integer 9.
In this problem, we will investigate this property for oth]]></description><category></category></item><item><title><![CDATA[字符串魔法(easy)]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/110950594</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/110950594</guid><author>qq_44527245</author><pubDate>Thu, 10 Dec 2020 12:38:26 +0800</pubDate><description><![CDATA[链接：https://ac.nowcoder.com/acm/contest/9680/B
来源：牛客网
白浅获得了一个仅由A和B组成的字符串。他可以至多使用一次魔法来改变字符串。 魔法的定义：选择一个字典序不递增的子串, 然后使得这个子串变成字典序不递减的子串，即变成形如AAA…AAABBB…BBB这样的字符串。 他想知道，在他至多使用一次魔法后，这个字符串能够出现的最长的字典序不递减的子串的长度为多少。
输入描述:
输入第一行包含一个整数n，代表字符串的长度
接下来一行给出一个长度为n的字符串(1≤n≤]]></description><category></category></item><item><title><![CDATA[全排列]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/110875767</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/110875767</guid><author>qq_44527245</author><pubDate>Tue, 08 Dec 2020 16:26:15 +0800</pubDate><description><![CDATA[链接：https://ac.nowcoder.com/acm/contest/9680/J
来源：牛客网
白浅经过一段时间对字符串的学习之后，学会了如何使用递归输出一个字符串的所有全排列（虽然 C++ 中有自带的函数），甚至学会了如何输出所有去重后的全排列（即“本质不同的全排列”）！她拿到了一个字符串 S，现在她将 S 的所有本质不同的全排列按字典序大小进行排序。假如总共有 k 个本质不同的字符串，则这些字符串按照字典序大小分别编号为 1.2.3…k，。白浅在这些字符串中任取了两个字符串 A B，现在她想知]]></description><category></category></item><item><title><![CDATA[爬塔(分组背包)]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/110783237</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/110783237</guid><author>qq_44527245</author><pubDate>Sun, 06 Dec 2020 22:32:01 +0800</pubDate><description><![CDATA[链接：https://ac.nowcoder.com/acm/contest/9680/F
来源：牛客网
高川最喜欢的游戏当属 Slay the Spire，这是一款爬塔游戏，你需要从一座塔的底部一直爬到顶部，在爬塔的过程中，塔的每一层都有许多的宝物等你来拿。
高川从塔的左侧开始攀爬，从底部爬到顶部，再从右侧从顶部逐步下到底部。塔总共有 n 层，每一层都有很多宝物从左到右排列。在左侧攀爬时，他只能从每层的最左边按顺序取宝物，在右侧下降时，他只能从每层的最右边按顺序取宝物。每个宝物都有一个价值，他最多拿 m ]]></description><category></category></item><item><title><![CDATA[Triangle]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/110481978</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/110481978</guid><author>qq_44527245</author><pubDate>Wed, 02 Dec 2020 15:28:54 +0800</pubDate><description><![CDATA[三分
#include &lt;stdio.h&gt;
#include &lt;math.h&gt;
#include &lt;algorithm&gt;
#include &lt;string.h&gt;
using namespace std;
double lz, lr, Lx, Ly, Rx, Ry, ans1, mid1, mid2, mid3, mid4, midx, midy, ans2, x1, x2, x3, p, q, y_1, y2, y3, l1, l2, l3;
doubl...]]></description><category></category></item><item><title><![CDATA[XOR-gun]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/110408472</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/110408472</guid><author>qq_44527245</author><pubDate>Mon, 30 Nov 2020 22:10:51 +0800</pubDate><description><![CDATA[Arkady owns a non-decreasing array a1,a2,…,an. You are jealous of its beauty and want to destroy this property. You have a so-called XOR-gun that you can use one or more times.
In one step you can select two consecutive elements of the array, let’s say x a]]></description><category></category></item><item><title><![CDATA[Cheerleaders（容斥原理）]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/110294556</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/110294556</guid><author>qq_44527245</author><pubDate>Sat, 28 Nov 2020 22:16:19 +0800</pubDate><description><![CDATA[In most professional sporting events, cheerleaders play a major role in entertaining the spectators. Their
roles are substantial during breaks and prior to start of play. The world cup soccer is no exception.
Usually the cheerleaders form a group and perfo]]></description><category></category></item><item><title><![CDATA[Piotr‘s Ants]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/109983603</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/109983603</guid><author>qq_44527245</author><pubDate>Mon, 23 Nov 2020 09:37:14 +0800</pubDate><description><![CDATA[Piotr likes playing with ants. He has n of them on a horizontal pole L cm long. Each ant is facing
either left or right and walks at a constant speed of 1 cm/s. When two ants bump into each other, they
both turn around (instantaneously) and start walking i]]></description><category></category></item><item><title><![CDATA[P2709 小B的询问（莫队）]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/109853404</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/109853404</guid><author>qq_44527245</author><pubDate>Fri, 20 Nov 2020 16:26:03 +0800</pubDate><description><![CDATA[#include &lt;stdio.h&gt;
#include &lt;algorithm&gt;
#include &lt;string.h&gt;
#include &lt;iostream&gt;
#include &lt;math.h&gt;
using namespace std;
int n, m, k, bbclock, a[1000001], cnt = 0, ans1 = 0;
struct node
{
    int l, r, id;
} b[1000001];
int c[..]]></description><category></category></item><item><title><![CDATA[K-th Number(主席树)]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/109823011</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/109823011</guid><author>qq_44527245</author><pubDate>Thu, 19 Nov 2020 19:43:41 +0800</pubDate><description><![CDATA[K-th Number
Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 81684		Accepted: 29562
Case Time Limit: 2000MS
Description
You are working for Macrohard company in data structures department. After failing your previous task about key insertion yo]]></description><category></category></item><item><title><![CDATA[Prefix Code(HASH+字典树)]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/109719108</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/109719108</guid><author>qq_44527245</author><pubDate>Mon, 16 Nov 2020 13:32:53 +0800</pubDate><description><![CDATA[对于给定的一串数字，看看是否有数字是其他数字的前缀。
HASH把每一个数字的前缀和数字本身都转换成对应的数字保存到一个数组，排序为非递减序列，之后用对每一个数字进行二分查找，如果能够找到并且找到的数字大于等于2.则输出No，当全部数字都被判断完，并且全部不是其他字符前缀则输出Yes，当字符形式输入，处理0为开始情况。
#include &lt;stdio.h&gt;
#include &lt;iostream&gt;
#include &lt;algorithm&gt;
#include &lt;st...]]></description><category></category></item><item><title><![CDATA[Color Graph]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/109659665</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/109659665</guid><author>qq_44527245</author><pubDate>Thu, 12 Nov 2020 21:56:39 +0800</pubDate><description><![CDATA[二分图的性质，不存在在奇环，一条边两端点在不同的集合，邻边不在同一边。
dfs枚举两个集合点的选择，选这个点在左边集合则在右边就不存在。
之后从所有的点的分配的情况中选一个合法边最多的情况。
#include &lt;iostream&gt;
#include &lt;stdio.h&gt;
#include &lt;string.h&gt;
#include &lt;algorithm&gt;
#include&lt;cmath&gt;
using namespace std;
int u[10001..]]></description><category></category></item><item><title><![CDATA[E. Team-Building]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/109644146</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/109644146</guid><author>qq_44527245</author><pubDate>Thu, 12 Nov 2020 15:18:27 +0800</pubDate><description><![CDATA[去掉不合适的小组（自身包含奇环）的情况下的所有组的两两组合数目减去不合适的组合。并根据可撤销并查集不停的考虑不同的两组之间的组合是否合适，没进行下一组之前便把之前在上一个两组之间加上的边都去掉。
如果两组之间连完边之后没有奇环的话就合适否则总数减一。
用并查集确定二分图是否含有奇环。
#include &lt;stdio.h&gt;
#include &lt;iostream&gt;
#include &lt;string.h&gt;
#include &lt;algorithm&gt;
using ...]]></description><category></category></item><item><title><![CDATA[D.Divide and Sum]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/109566201</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/109566201</guid><author>qq_44527245</author><pubDate>Sun, 08 Nov 2020 22:32:20 +0800</pubDate><description><![CDATA[对a的元素进行排序。
将a的元素放到p,q两个队列中，对每一个元素对f的贡献值单独考虑。
对于f（xi,yi）=|xi-yi|，当中大的数做正贡献，小的做负贡献。
对于ai个元素假设前面有k个元素在p队列，则i-1-k个元素在q
如果要ai有正贡献，则k&gt;=n-(i-1-k);否则负贡献。
如果在q队列中对于ai，设有k个在q队列中，则i-1-k个在q队列中，要做正贡献，则k&gt;=n-(i-1-k)‘
两个的结果的是i&gt;n做正贡献，否则负贡献。
所以对于所有的排列组合中，i&gt;n的...]]></description><category></category></item><item><title><![CDATA[C - Shuffle Permutation(并查集+排列组合)]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/109566158</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/109566158</guid><author>qq_44527245</author><pubDate>Sun, 08 Nov 2020 22:18:50 +0800</pubDate><description><![CDATA[从行和列两方面考虑问题。
1.从行，暴力遍历从第一行到最后一行，每两行的以元素，两两相加把每一个元素相加都小于k的两行都放到一个并查集，这里类似朋友的朋友也是朋友，如A与B符合条件，A与C符合条件，则A，B，C都在一个并查集，因为这里要利用通过变换能够得到多少不同种矩阵。只要在一个矩阵里就可以用排列组合的方式求有多少种分布方式。每一行中的不同并查集之间的分布种数通过相乘得到行通过变换可以得到多少种排列矩阵的方式。
2.从列，与行相同。
3.最后行的方式与列的方式相乘得到结果。
#includ...]]></description><category></category></item><item><title><![CDATA[C. Division]]></title><link>https://blog.youkuaiyun.com/qq_44527245/article/details/109480203</link><guid>https://blog.youkuaiyun.com/qq_44527245/article/details/109480203</guid><author>qq_44527245</author><pubDate>Tue, 03 Nov 2020 21:55:58 +0800</pubDate><description><![CDATA[#include &lt;stdio.h&gt;
#define max(a, b) (a &gt; b ? a : b)
long long int ans = 1;
void f(long long int p, long long int q,long long int x)
{
    while (p % q == 0)
    {
        p /= x;
    }
    ans = max(ans, p);
}
int main()
{
#ifdef ONLINE_JUDGE
#..]]></description><category></category></item></channel></rss>