<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[l2ice的专栏]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/l2ice</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; l2ice]]></copyright><item><title><![CDATA[poj 2的n次方 http://poj.grids.cn/problem/2809/]]></title><link>https://blog.youkuaiyun.com/l2ice/article/details/5556629</link><guid>https://blog.youkuaiyun.com/l2ice/article/details/5556629</guid><author>l2ice</author><pubDate>Tue, 04 May 2010 17:09:00 +0800</pubDate><description><![CDATA[稍有常识的人都会看出，做完了前面那道题后，这道题还会做不出来么？#include#includeusing namespace std;int a1[40]={0};int a2[40]={0};void multiply(int a[],int b[]){	int c[40]={0};	for(int i=0;i<40;i++)	{		int e=0;]]></description><category></category></item><item><title><![CDATA[麦森数 poj http://poj.grids.cn/solution/1154879/]]></title><link>https://blog.youkuaiyun.com/l2ice/article/details/5556246</link><guid>https://blog.youkuaiyun.com/l2ice/article/details/5556246</guid><author>l2ice</author><pubDate>Tue, 04 May 2010 15:52:00 +0800</pubDate><description><![CDATA[题目：http://poj.grids.cn/solution/1154879/个人认为这道题 在高精度计算中应该是算有难度的，这道题用了很多技巧1首先2^p计算要用到一个技巧：首先把p用二进制表示p=a0*2^0+a1*2^1+……+an*2^n-1+2^n;那么2^p=2^a0*2^2a1*2^4a2*……*2^2^n;用算法表示就是 while(P) {  i]]></description><category></category></item><item><title><![CDATA[大整数除法poj http://poj.grids.cn/problem/2737/]]></title><link>https://blog.youkuaiyun.com/l2ice/article/details/5548380</link><guid>https://blog.youkuaiyun.com/l2ice/article/details/5548380</guid><author>l2ice</author><pubDate>Sat, 01 May 2010 13:11:00 +0800</pubDate><description><![CDATA[ http://poj.grids.cn/problem/2737/这道题判断数的大小部分。。由于想得太简单了。。所以一开始就写错了最后还是ac了 题目基本简单。。但复杂。。#includeusing namespace std;int len1,len2;bool sub(int a1[],int a2[],int len){	int i=len-1;	i]]></description><category></category></item><item><title><![CDATA[poj 大整数加法 http://poj.grids.cn/problem/2981/]]></title><link>https://blog.youkuaiyun.com/l2ice/article/details/5545976</link><guid>https://blog.youkuaiyun.com/l2ice/article/details/5545976</guid><author>l2ice</author><pubDate>Fri, 30 Apr 2010 13:26:00 +0800</pubDate><description><![CDATA[http://poj.grids.cn/problem/2981/这道题要注意 00情况；开始时wa就因为这个了#include#includeint main(){	char a[201]={0},b[201]={0};	scanf("%s %s",a,b);	int lena=strlen(a);	int lenb=strlen(b);	in]]></description><category></category></item><item><title><![CDATA[poj循环数 ]]></title><link>https://blog.youkuaiyun.com/l2ice/article/details/5544525</link><guid>https://blog.youkuaiyun.com/l2ice/article/details/5544525</guid><author>l2ice</author><pubDate>Fri, 30 Apr 2010 00:17:00 +0800</pubDate><description><![CDATA[http://poj.grids.cn/problem/2952/题目较为简单 我就不解释了 #includeusing namespace std;int len=0;bool add(int a[],char b[],int v2[],int v1[]){	for(int i=len-1;i>=0;i--)	{		a[i]+=b[i]-0;		if(]]></description><category></category></item><item><title><![CDATA[poj武林 http://poj.grids.cn/problem/2785/]]></title><link>https://blog.youkuaiyun.com/l2ice/article/details/5544300</link><guid>https://blog.youkuaiyun.com/l2ice/article/details/5544300</guid><author>l2ice</author><pubDate>Thu, 29 Apr 2010 22:46:00 +0800</pubDate><description><![CDATA[这个题不是很难，但如果时间复杂度方面想太多的话，会弄得十分麻烦 这个代码也不完全是自己写的，开始是自己写，用了太多指针的东西，结果re了很多次。最后还是扫了下别人的代码，然后自己重编了一下；但因为为了追求代码精简。下面这 个地方想得太简单结果弄错if(a[i].i==12||a[i].j==12)a[i].dir=1;else if(a[i].i==1||a[i].j==1)]]></description><category></category></item></channel></rss>