<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[dukig的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/dukig</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; dukig]]></copyright><item><title><![CDATA[分治]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/94624344</link><guid>https://blog.youkuaiyun.com/dukig/article/details/94624344</guid><author>dukig</author><pubDate>Thu, 04 Jul 2019 15:50:11 +0800</pubDate><description><![CDATA[poj 1854

求一个字符串相邻字母经过多少次交换后可以让这个字符串变成回文串。

解法：对首尾字母分治


#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;cstdio&gt;
#include&lt;cstring&gt;
#include&lt;cstdlib&gt;
using namespace std;
...]]></description><category></category></item><item><title><![CDATA[洛谷挑战LV3--数论]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89854044</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89854044</guid><author>dukig</author><pubDate>Sun, 05 May 2019 20:00:12 +0800</pubDate><description><![CDATA[P2154

java自带大数据处理


import java.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Scanner;

public class Main {
     static public void main(String args[]) {
    	 Scanner...]]></description><category></category></item><item><title><![CDATA[容斥原理专题]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89714089</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89714089</guid><author>dukig</author><pubDate>Tue, 30 Apr 2019 17:59:56 +0800</pubDate><description><![CDATA[原理在第三行最后一列

hdoj1465

设n个装错，n-1,n-2..


#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;cstdio&gt;
#include&lt;cstring&gt;
using namespace std;

int f[21];

void init(){
	for(int i = 2;...]]></description><category></category></item><item><title><![CDATA[2017 沈阳区域赛]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89481202</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89481202</guid><author>dukig</author><pubDate>Sun, 28 Apr 2019 12:01:04 +0800</pubDate><description><![CDATA[F （思维）

题目大意：有n个兔子排成一列在河边玩游戏，每个兔子占有一个不同的数字，游戏规则是最外边的兔子，可以在其他任意两只兔子之间移动，求最多可以移动的次数，（一个兔子一个坑）。

纯思维 签到题


#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;cstdio&gt;
#include&lt;cstring&gt...]]></description><category></category></item><item><title><![CDATA[2017 香港]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89575832</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89575832</guid><author>dukig</author><pubDate>Sun, 28 Apr 2019 12:00:54 +0800</pubDate><description><![CDATA[E（最大化最小值）

模板题


#include &lt;cstdio&gt;
#include &lt;iostream&gt;
#include &lt;algorithm&gt;
#include &lt;cstring&gt;
#include &lt;cmath&gt;
#define INF 0x3f3f3f3f
using namespace std;
const int MAXN...]]></description><category></category></item><item><title><![CDATA[17青岛区域赛]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89477640</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89477640</guid><author>dukig</author><pubDate>Sun, 28 Apr 2019 12:00:41 +0800</pubDate><description><![CDATA[链接：https://nanti.jisuanke.com/acm?kw=ACM-ICPC%202017%20Asia%20Qingdao

B（模拟）

像#一样3个一个字母就输出那个字母


#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;cstdio&gt;
#include&lt;cstring&gt;
usin...]]></description><category></category></item><item><title><![CDATA[2013 Asia Regional Changchun]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89601902</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89601902</guid><author>dukig</author><pubDate>Sun, 28 Apr 2019 12:00:34 +0800</pubDate><description><![CDATA[A

非常简单的签到题


#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;cstdio&gt;
#include&lt;cstring&gt;
using namespace std;

int main(){
	int t,n,m;
	string a;
	scanf("%d",&amp;t);
	while(t--...]]></description><category></category></item><item><title><![CDATA[2018多校 第一场]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89459691</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89459691</guid><author>dukig</author><pubDate>Sat, 27 Apr 2019 15:09:34 +0800</pubDate><description><![CDATA[A

给出一个数n，找到3个数a,b,c，a,b,c均能整除n，使得n = a+b+c且a*b*c最大，找不到a b c就输出-1

令 x = n/a y = n/b z = n/c 公式同时除以n---&gt;1 = 1/x+1/y+1/z -- &gt;3个解 （3，3，3）（2，4，4），（2，3，6）

当n%3 == 0--&gt; x*y*z = n^3/27

n%4 == 0 -...]]></description><category></category></item><item><title><![CDATA[UVA-11827 （输入坑）]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89576395</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89576395</guid><author>dukig</author><pubDate>Fri, 26 Apr 2019 20:42:46 +0800</pubDate><description><![CDATA[链接：https://cn.vjudge.net/contest/276155#problem/V

输入可以学学


#include &lt;iostream&gt;
#include &lt;cstdlib&gt;
#include &lt;cstdio&gt;
 
using namespace std;
 
int data[101];
 
int gcd(int a, int b)
{...]]></description><category></category></item><item><title><![CDATA[uva11752 （素数）]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89575974</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89575974</guid><author>dukig</author><pubDate>Fri, 26 Apr 2019 20:13:49 +0800</pubDate><description><![CDATA[题意：求0到上限的是至少2个数的平方的数。

那个数一定是和数


#define _CRT_SECURE_NO_WARNINGS 
#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;string&gt;
#include&lt;sstream&gt;
#include&lt;set&gt;
#include&lt;vec...]]></description><category></category></item><item><title><![CDATA[UVA - 11426 （欧拉函数）]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89526122</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89526122</guid><author>dukig</author><pubDate>Thu, 25 Apr 2019 21:01:29 +0800</pubDate><description><![CDATA[链接：https://vjudge.net/contest/276155#problem/O

题意：给出n,求gcd(i,j)得总和(i重1到n-1，j重i+1到n-1)

用欧拉函数求出数量 很巧妙


#include &lt;cstdio&gt;
#include &lt;cstring&gt;
 
using namespace std;
 
const int MAXN = 40000...]]></description><category></category></item><item><title><![CDATA[LightOJ - 1138 （求n的阶乘末尾0的个数）]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89524729</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89524729</guid><author>dukig</author><pubDate>Thu, 25 Apr 2019 20:08:48 +0800</pubDate><description><![CDATA[https://vjudge.net/contest/276155#problem/N

因为只有与5相乘才会得0，所以只用求5的个数即可。


#include&lt;iostream&gt;
#include&lt;cstdlib&gt;
#include&lt;cstdio&gt;
#include&lt;algorithm&gt;
#include&lt;cstring&gt;
#incl...]]></description><category></category></item><item><title><![CDATA[LightOJ - 1197（大整数素数塞）记住]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89459312</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89459312</guid><author>dukig</author><pubDate>Mon, 22 Apr 2019 19:47:05 +0800</pubDate><description><![CDATA[https://vjudge.net/contest/276155#problem/M


#include&lt;cstdio&gt;
#include&lt;cmath&gt;
#include&lt;algorithm&gt;
#include&lt;cstring&gt;
#include&lt;iostream&gt;
using namespace std;
typedef long ...]]></description><category></category></item><item><title><![CDATA[LightOJ - 1213（组合数学 快速幂）]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89458730</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89458730</guid><author>dukig</author><pubDate>Mon, 22 Apr 2019 19:15:57 +0800</pubDate><description><![CDATA[https://vjudge.net/contest/276155#problem/L

给出n,k,mod 和数量为n得int数组a[i]，求k层嵌套得a数组相加后膜mod的值。

排列组合求得每个数的出现次数为(n^(k-1))*k,再用快速幂求结果。


#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;cstdio...]]></description><category></category></item><item><title><![CDATA[知识点]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89439200</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89439200</guid><author>dukig</author><pubDate>Sun, 21 Apr 2019 20:50:51 +0800</pubDate><description><![CDATA[求1~n得约数和：https://ac.nowcoder.com/acm/problem/14682

海伦海勒


for (t=1; t&lt;=100000; t++)   //求面积为整数的
	{
		int delta=3*(t*t-4);
		int pd=floor(sqrt(delta));
		if (pd*pd!=delta) continue;
		delta=pd;
		i...]]></description><category></category></item><item><title><![CDATA[LightOJ - 1214（大数取模）]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89438471</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89438471</guid><author>dukig</author><pubDate>Sun, 21 Apr 2019 20:16:40 +0800</pubDate><description><![CDATA[链接：https://cn.vjudge.net/contest/276155#problem/K

求a能否整除b。

将大数分解为一个个数，例如123 = 1*10^2 + 2*10^1 + 3*10^0,逐个膜。


#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;cstdio&gt;
#include&lt;cs...]]></description><category></category></item><item><title><![CDATA[LightOJ - 1220 (唯一分解定理)]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89438073</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89438073</guid><author>dukig</author><pubDate>Sun, 21 Apr 2019 19:55:01 +0800</pubDate><description><![CDATA[题意：给定n,n = b^p,b与p均未给出，求最大的p。

用唯一分解定理分解n，p就是n所有因子得幂得最大公约数。如 12 = 2^2*3,p = gcd(2,1) = 1，n为负数时p必须处理为奇数，因为偶数时n就是正数了。


#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;cstdio&gt;
#include...]]></description><category></category></item><item><title><![CDATA[LightOJ - 1234（欧拉常数 模板记）]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89437346</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89437346</guid><author>dukig</author><pubDate>Sun, 21 Apr 2019 19:09:37 +0800</pubDate><description><![CDATA[链接：https://cn.vjudge.net/contest/276155#problem/I

1+1/2+1/3... = log(1+n) + r(欧拉常数),需要大精度时减去1.0/(2*n);

r =0.57721566490153286060651209


#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&...]]></description><category></category></item><item><title><![CDATA[湘潭ccpc2018]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89433761</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89433761</guid><author>dukig</author><pubDate>Sun, 21 Apr 2019 15:35:07 +0800</pubDate><description><![CDATA[链接：https://cn.vjudge.net/contest/296332#overview

A.

阅读理解题。找出引用数大于h的至少有h篇的论文那个最大的h数。（我晕了）

比如 第二个样例 2 1(下标0）2(下标1） 3(下标2）

其中下标表示引用数，而ai表示论文的数论，3(下标2）表示引用了2页的论文有3篇。

这里引用数大于2的为3篇，引用数大于1的为5篇，引用数大于0的...]]></description><category></category></item><item><title><![CDATA[牛客寒假3部分]]></title><link>https://blog.youkuaiyun.com/dukig/article/details/89432876</link><guid>https://blog.youkuaiyun.com/dukig/article/details/89432876</guid><author>dukig</author><pubDate>Sun, 21 Apr 2019 14:46:47 +0800</pubDate><description><![CDATA[链接：https://ac.nowcoder.com/acm/contest/329#question



I.

怎样都行，要么把它变成全1，要么把它变为全0.


#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;cstdio&gt;
#include&lt;cstring&gt;
using namespace ...]]></description><category></category></item></channel></rss>