<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[二十五之前我可能会秃头]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/acmmiao</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; acmmiao]]></copyright><item><title><![CDATA[UVA 11292 Dragon of Loowater]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/52887237</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/52887237</guid><author>acmmiao</author><pubDate>Fri, 21 Oct 2016 21:37:45 +0800</pubDate><description><![CDATA[题解: 贪心.. 恶龙有n个头 有m个猎人
n     m
x(共n行   x表示砍断相应头所需的值)
x
....
x
y(共m行  y表示猎人的砍龙头的能力也是雇佣金的数目)c
y
...
y
排序+贪心(水题)


AC代码:
#include 
#include 
#include 
#include 
#include 
#include 
#include ]]></description><category></category></item><item><title><![CDATA[Codeforces Round #376 (Div. 2)]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/52864723</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/52864723</guid><author>acmmiao</author><pubDate>Wed, 19 Oct 2016 23:13:26 +0800</pubDate><description><![CDATA[A - Night at the Museum
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d
 & %I64u
Submit Status Practice CodeForces
 731A





Description

Grigoriy, like the hero of]]></description><category></category></item><item><title><![CDATA[hdu 1753 大明A+B]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51931514</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51931514</guid><author>acmmiao</author><pubDate>Sun, 17 Jul 2016 11:42:57 +0800</pubDate><description><![CDATA[Description

话说，经过了漫长的一个多月，小明已经成长了许多，所以他改了一个名字叫“大明”。 
这时他已经不是那个只会做100以内加法的那个“小明”了，现在他甚至会任意长度的正小数的加法。 

现在，给你两个正的小数A和B，你的任务是代表大明计算出A+B的值。 

 




Input

本题目包含多组测试数据，请处理到文件结束。 
每一组测试数据在一行]]></description><category></category></item><item><title><![CDATA[Poj 3304 Segments]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51820171</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51820171</guid><author>acmmiao</author><pubDate>Mon, 04 Jul 2016 13:08:17 +0800</pubDate><description><![CDATA[Description


Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at]]></description><category></category></item><item><title><![CDATA[Poj 1654 Area]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51818575</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51818575</guid><author>acmmiao</author><pubDate>Mon, 04 Jul 2016 09:25:07 +0800</pubDate><description><![CDATA[Description


You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From this vertex, you may go step by step ]]></description><category></category></item><item><title><![CDATA[Poj 1269 Intersecting Lines]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51818499</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51818499</guid><author>acmmiao</author><pubDate>Mon, 04 Jul 2016 09:14:18 +0800</pubDate><description><![CDATA[Description


We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are paral]]></description><category></category></item><item><title><![CDATA[取(2堆)石子游戏]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51804623</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51804623</guid><author>acmmiao</author><pubDate>Fri, 01 Jul 2016 22:53:30 +0800</pubDate><description><![CDATA[Description

有两堆石子，数量任意，可以不同。游戏开始由两个人轮流取石子。游戏规定，每次有两种不同的取法，一是可以在任意的一堆中取走任意多的石子；二是可以在两堆中同时取走相同数量的石子。最后把石子全部取完者为胜者。现在给出初始的两堆石子的数目，如果轮到你先取，假设双方都采取最好的策略，问最后你是胜者还是败者。如果你胜，你第1次怎样取子? 

 




Input]]></description><category></category></item><item><title><![CDATA[HDU 2149 Public Sale（巴什博奕）]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51804612</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51804612</guid><author>acmmiao</author><pubDate>Fri, 01 Jul 2016 22:48:23 +0800</pubDate><description><![CDATA[Description

虽然不想，但是现实总归是现实，Lele始终没有逃过退学的命运，因为他没有拿到奖学金。现在等待他的，就是像FarmJohn一样的农田生涯。 

要种田得有田才行，Lele听说街上正在举行一场别开生面的拍卖会，拍卖的物品正好就是一块20亩的田地。于是，Lele带上他的全部积蓄，冲往拍卖会。 

后来发现，整个拍卖会只有Lele和他的死对头Yueyue。 

通]]></description><category></category></item><item><title><![CDATA[Poj 1067 取石子游戏]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51804599</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51804599</guid><author>acmmiao</author><pubDate>Fri, 01 Jul 2016 22:45:54 +0800</pubDate><description><![CDATA[Description

有两堆石子，数量任意，可以不同。游戏开始由两个人轮流取石子。游戏规定，每次有两种不同的取法，一是可以在任意的一堆中取走任意多的石子；二是可以在两堆中同时取走相同数量的石子。最后把石子全部取完者为胜者。现在给出初始的两堆石子的数目，如果轮到你先取，假设双方都采取最好的策略，问最后你是胜者还是败者。
 




Input

输入包含若干行，表示若干种石子]]></description><category></category></item><item><title><![CDATA[HDU 1222 Wolf and Rabbit（gcd）]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51771841</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51771841</guid><author>acmmiao</author><pubDate>Mon, 27 Jun 2016 22:30:28 +0800</pubDate><description><![CDATA[Description

There is a hill with n holes around. The holes are signed from 0 to n-1. 




A rabbit must hide in one of the holes. A wolf searches the rabbit in anticlockwise order. The first h]]></description><category></category></item><item><title><![CDATA[POJ 2262 Goldbach's Conjecture(素数)]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51771825</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51771825</guid><author>acmmiao</author><pubDate>Mon, 27 Jun 2016 22:28:51 +0800</pubDate><description><![CDATA[Description


In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture: 
Every even number greater than 4 can be 
wri]]></description><category></category></item><item><title><![CDATA[A/B（乘法逆元）]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51771808</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51771808</guid><author>acmmiao</author><pubDate>Mon, 27 Jun 2016 22:25:08 +0800</pubDate><description><![CDATA[A/B




Description

要求(A/B)%9973，但由于A很大，我们只给出n(n=A%9973)(我们给定的A必能被B整除，且gcd(B,9973) = 1)。
 




Input

数据的第一行是一个T，表示有T组数据。 
每组数据有两个数n(0 
 




Output

对应每组数据输出(A/B)%9973。
 
]]></description><category></category></item><item><title><![CDATA[欧几里德与扩展欧几里德算法]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51771128</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51771128</guid><author>acmmiao</author><pubDate>Mon, 27 Jun 2016 21:58:21 +0800</pubDate><description><![CDATA[欧几里德算法（辗转相除法） 可用于计算两个整数a,b的公约数和公倍数
定理：gcd(a, b) = gcd(b , a%b)


int gcd(int a,int b)
{
  return b ? gcd(b,a%b):a;
}
最小公倍数是：  a*b/gcd(a,b) 不过我们为了防止a*b的溢出 采取先除后乘
a/gcd(a,b)*b
扩展欧几里德算法
基本算法：对于不完]]></description><category></category></item><item><title><![CDATA[COJ 1010  [001]质因数分解 [EASY]]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51750358</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51750358</guid><author>acmmiao</author><pubDate>Fri, 24 Jun 2016 10:09:43 +0800</pubDate><description><![CDATA[[001]质因数分解 [EASY]
Time Limit: 1000 ms     Memory Limit: 65536 KB
Total Submit: 1891     Accepted: 582 


Description

找出输入整数的所有质因数（包括重复质因数），并按从小到大的顺序依次输出。




Input

输入一组待分解整数，每个整数k占一行。]]></description><category></category></item><item><title><![CDATA[HDU 1248 寒冰王座 (完全背包)]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51494571</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51494571</guid><author>acmmiao</author><pubDate>Tue, 24 May 2016 23:40:53 +0800</pubDate><description><![CDATA[Description

不死族的巫妖王发工资拉,死亡骑士拿到一张N元的钞票(记住,只有一张钞票),为了防止自己在战斗中频繁的死掉,他决定给自己买一些道具,于是他来到了地精商店前. 

死亡骑士:"我要买道具!" 

地精商人:"我们这里有三种道具,血瓶150块一个,魔法药200块一个,无敌药水350块一个." 

死亡骑士:"好的,给我一个血瓶." 

说完他掏出那张N元的大]]></description><category></category></item><item><title><![CDATA[POJ 3624 Charm Bracelet (01背包)]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51494559</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51494559</guid><author>acmmiao</author><pubDate>Tue, 24 May 2016 23:38:42 +0800</pubDate><description><![CDATA[Description


Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N(1 ≤ N ≤ 3,402) available charms. Each]]></description><category></category></item><item><title><![CDATA[HDU 2602 Bone Collector（01背包）]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51494551</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51494551</guid><author>acmmiao</author><pubDate>Tue, 24 May 2016 23:37:15 +0800</pubDate><description><![CDATA[Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave … 
Th]]></description><category></category></item><item><title><![CDATA[5-38 数列求和-加强版   (20分)]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51494534</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51494534</guid><author>acmmiao</author><pubDate>Tue, 24 May 2016 23:33:18 +0800</pubDate><description><![CDATA[#include 
#include 
#include 
using namespace std;

int a[10000020]= {0};
int main()
{
    int A,N,b,i=0;
    scanf("%d%d",&A,&N);
    if(N==0) printf("%d",N);
    else
    {
        int m=N;
        ]]></description><category></category></item><item><title><![CDATA[5-37 整数分解为若干项之和   (20分)]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51494532</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51494532</guid><author>acmmiao</author><pubDate>Tue, 24 May 2016 23:32:49 +0800</pubDate><description><![CDATA[#include
int s[100];//拆分结果保存在这个数组里
int top;//记录个数
int total, n;//累加数和所求数
int k;
void dfs(int index)
{
        int i;
        if (total == n){
                printf("%d=", n);
                for (i =]]></description><category></category></item><item><title><![CDATA[5-36 复数四则运算   (15分)]]></title><link>https://blog.youkuaiyun.com/acmmiao/article/details/51494529</link><guid>https://blog.youkuaiyun.com/acmmiao/article/details/51494529</guid><author>acmmiao</author><pubDate>Tue, 24 May 2016 23:32:21 +0800</pubDate><description><![CDATA[#include
#include
#include
#include
using namespace std;
int main()
{
	double a1,i1,a2,i2;
	cin>>a1>>i1>>a2>>i2;
	double jiashi,jiaxu;
	double jianshi,jianxu;
	double chengshi,chengxu;
	double chushi,]]></description><category></category></item></channel></rss>