<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[sevanco的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/sevanco</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; sevanco]]></copyright><item><title><![CDATA[阿里笔试题二]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/70832690</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/70832690</guid><author>sevanco</author><pubDate>Wed, 26 Apr 2017 21:17:57 +0800</pubDate><description><![CDATA[#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>/** 请完成下面这个函数，实现题目要求的功能 **/
 /** 当然，你也可以不按照这个模板来作答，完全按照自己的想法来 ^-]]></description><category></category></item><item><title><![CDATA[阿里笔试题一]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/70832682</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/70832682</guid><author>sevanco</author><pubDate>Wed, 26 Apr 2017 21:17:14 +0800</pubDate><description><![CDATA[#include <iostream>
#include <vector>
#include <numeric>
#include <limits>
#include <math.h>
using namespace std;/** 请完成下面这个函数，实现题目要求的功能 **/
 /** 当然，你也可以不按照这个模板来作答，完全按照自己的想法来 ^-^  **/int prime(int num_]]></description><category></category></item><item><title><![CDATA[单链表的递归逆置]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/68940645</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/68940645</guid><author>sevanco</author><pubDate>Sat, 01 Apr 2017 08:14:36 +0800</pubDate><description><![CDATA[#include <stdio.h>
#include <stdlib.h>
#define M 20
struct num
{
    int a;
    struct num *next;
};
struct num *reverse(struct num *head)
{
    struct num *p;
    p = head;
    if (p->next == NULL)]]></description><category></category></item><item><title><![CDATA[单项链表排序]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/68923987</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/68923987</guid><author>sevanco</author><pubDate>Fri, 31 Mar 2017 13:15:37 +0800</pubDate><description><![CDATA[今天复习单项链表。自己编了一上午，蛋疼。。。#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define M 2
struct student
{
    char name[M];
    double score;
    struct student *next;
};
struct s]]></description><category></category></item><item><title><![CDATA[结构体对齐问题]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/68921148</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/68921148</guid><author>sevanco</author><pubDate>Fri, 31 Mar 2017 08:30:10 +0800</pubDate><description><![CDATA[昨天华为面试，问到了关于结构体对齐的问题。我懵逼了，结构体对齐是什么梗？一开始还以为是和资源受限的设备（单片机啊之类的嵌入式设备）开发有关。 
今天下了下资料，发现原理是真的简单，当时多考虑一会应该能想出来的。struct {
    char a;      //1byte
    int b;       //4byte
    char c[2]    //2byte
    double d]]></description><category></category></item><item><title><![CDATA[字符串去重]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/67644313</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/67644313</guid><author>sevanco</author><pubDate>Tue, 28 Mar 2017 20:34:43 +0800</pubDate><description><![CDATA[#include <stdio.h>
#include <string.h>
#define M 500void fun(char *a, void (*p)(char *, void (*q)(char *, char *)), void  (*q)(char *, char *))
{
    char *t;
    for (t = a + 1; *t != '\0'; ++t)
    {]]></description><category></category></item><item><title><![CDATA[将double类型二维数组的每一行同除以该行上绝对值最大的元素]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/66971615</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/66971615</guid><author>sevanco</author><pubDate>Mon, 27 Mar 2017 14:49:22 +0800</pubDate><description><![CDATA[#include <stdio.h>
#include <string.h>
#include <math.h>
#define M 2
#define N 2
void fun(int b, double a[][N], double t)
{
    for (int i = 0; i < N; ++i)
    {
        // printf("asdas\n");
        /]]></description><category></category></item><item><title><![CDATA[输入5个字符串,输出最长的字符串.]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/66970587</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/66970587</guid><author>sevanco</author><pubDate>Mon, 27 Mar 2017 13:14:10 +0800</pubDate><description><![CDATA[#include <stdio.h>
#include <string.h>
#define M 5
const char *com(const char **a)
{
    int t = 0;
    for (int i = 0; i < M - 1; ++i)
    {
        if (strlen(*(a + t)) < strlen(*(a + i + 1)))]]></description><category></category></item><item><title><![CDATA[函数指针练习]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/66970338</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/66970338</guid><author>sevanco</author><pubDate>Mon, 27 Mar 2017 12:24:04 +0800</pubDate><description><![CDATA[#include <stdio.h>
#include <string.h>
#define M 5
void exc(int *a,int *(*p)(int *), int *(*q)(int *))
{
    int c, *m, *n; 
    m = p(a);
    n = q(a);
    c = *m;
    *m = *n;
    *n = c;
}
int *max(]]></description><category></category></item><item><title><![CDATA[deprecated conversion from string constant to ‘char*’ [-Wwrite-strings]]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/66967806</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/66967806</guid><author>sevanco</author><pubDate>Mon, 27 Mar 2017 08:56:39 +0800</pubDate><description><![CDATA[今天在复习C语言指针多字符串的时候遇到的一个小问题，我使用的是g++编译器，所以在运行代码的时候报错了。 
代码：#include <stdio.h>
#include <string.h>
#define M 200
int main(int argc, char const *argv[])
{
    char str[M] = "asdasd";
    static char *name[]]></description><category></category></item><item><title><![CDATA[C语言的行指针]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/66476555</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/66476555</guid><author>sevanco</author><pubDate>Sun, 26 Mar 2017 16:29:11 +0800</pubDate><description><![CDATA[今天复习指针的时候看到书上讲二维数组中的两类指针：
一类是元素指针
一类是行指针
元素指针指向一个元素，行指针指向整个行。（这里比较难以理解的是行指针）行指针可以转化为元素指针，方法是在前面加一个”∗*”,指向改行的第一个元素。也就是说加上了”∗*”之后，它还是一个地址。代码#include <stdio.h>
#include <string.h>
#define M 2
#define N 2]]></description><category></category></item><item><title><![CDATA[卜踆哉数对]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/66472618</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/66472618</guid><author>sevanco</author><pubDate>Sun, 26 Mar 2017 10:50:04 +0800</pubDate><description><![CDATA[阿里笔试题—卜踆哉对数今天下午做完网易笔试题，然后5点准备做腾讯的面试题，然后发现只是模拟试卷，汗！-_-!!言归正传，晚上阿里试题题目是卜踆哉数对，网上也没有相关资料。（怕不是阿里程序猿现编的数学数对，雾2333333）最后虽然好像找到了一条路径，但还是没时间去实现了。（40分钟真是短~~~）本文介绍一下我的思路（有可能并不好hhhhh）目录阿里笔试题卜踆哉对数目录
题目
过程
初来乍到
RSA内]]></description><category></category></item><item><title><![CDATA[数组的指针操作]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/65935249</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/65935249</guid><author>sevanco</author><pubDate>Sat, 25 Mar 2017 10:57:56 +0800</pubDate><description><![CDATA[#include 
int sum(int *b, int n)
{
int m = 0, *p;
for (p = b; p 
{
printf("m = %d\n", m);
printf("p = %d\n", *p);
m += *p;
}
return m;


}
int main(int argc, char const *argv[])
{
int ]]></description><category></category></item><item><title><![CDATA[矩阵乘法的C语言实现]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/65627517</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/65627517</guid><author>sevanco</author><pubDate>Fri, 24 Mar 2017 10:20:53 +0800</pubDate><description><![CDATA[#include 
#include 
#include 
#define M 2
#define N 3
int main(int argc, char const *argv[])
{
int a[M][N] = {{1, 2, 3}, {4, 5, 6}};
int b[N][M] = {{1, 2}, {3, 4}, {5, 6}};
int re = 0;
int c]]></description><category></category></item><item><title><![CDATA[约瑟夫环的递归思想解]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/64907398</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/64907398</guid><author>sevanco</author><pubDate>Wed, 22 Mar 2017 12:50:26 +0800</pubDate><description><![CDATA[原文地址：http://blog.youkuaiyun.com/u014613043/article/details/50905475
今天做约瑟夫环的题目，对于递归解法不太了解。在网上找了一篇比较好理解的文章。


本文为学习《剑指offer》的记录。因其原理在原作者博客上找不到，所以，只能自己编写记录，如有不当之处，欢迎指正。

题目描述： 
n个数，编号为 0 ， 1， ……， n-1 排]]></description><category></category></item><item><title><![CDATA[年份信息查询程序]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/63252673</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/63252673</guid><author>sevanco</author><pubDate>Sat, 18 Mar 2017 12:03:46 +0800</pubDate><description><![CDATA[#include 
#include 
int days[] = {31,28,31,30,31,30,31,31,30,31,30,31};
int days2[] = {31,29,31,30,31,30,31,31,30,31,30,31};
int year = 0, mon = 0, day = 0;
int choice = 0;
int check();
int run]]></description><category></category></item><item><title><![CDATA[hanoit递归问题]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/62230133</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/62230133</guid><author>sevanco</author><pubDate>Wed, 15 Mar 2017 16:52:37 +0800</pubDate><description><![CDATA[#include 
void move(char a, char b)
{
printf("%c --> %c\n", a, b);
}
void hanoit(int n, char a, char b, char c)
{
if (n == 1)
{
move(a, c);
}
else
{
hanoit(n - 1, a, c, b);
move(a, c);
]]></description><category></category></item><item><title><![CDATA[C语言中关于pow()函数的问题]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/61917355</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/61917355</guid><author>sevanco</author><pubDate>Mon, 13 Mar 2017 14:40:11 +0800</pubDate><description><![CDATA[今天在码代码时发现pow()和基本的变量乘法似乎有出入。
代码：
#include 
#include 
int main(int argc, char const *argv[])
{
int t = 0, flag = 0;
int m = -39, n = 40;
for (int i = m; i 
{
printf("%d\n", pow(-39,2));
prin]]></description><category></category></item><item><title><![CDATA[关于编程时变量类型转换问题]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/61916161</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/61916161</guid><author>sevanco</author><pubDate>Mon, 13 Mar 2017 13:24:49 +0800</pubDate><description><![CDATA[今天在做证收敛的代码时，发现值永远是0.
代码：
#include 
int main(int argc, char const *argv[])
{
int n = 1/*, sign = 0*/;
double sign = -1;
double t = 0, m = 0;
while(n 
{
sign = -1 * sign;
t += sign * 1 / n;
]]></description><category></category></item><item><title><![CDATA[关于sublime 编译C与gc++编译C的一个问题]]></title><link>https://blog.youkuaiyun.com/sevanco/article/details/61915398</link><guid>https://blog.youkuaiyun.com/sevanco/article/details/61915398</guid><author>sevanco</author><pubDate>Mon, 13 Mar 2017 11:48:04 +0800</pubDate><description><![CDATA[今天做水仙花数的时候谢了一段代码，但是发现sublime下运行结果与gc++6.0编译结果不一致，记录以究。
#include 
#include 
int main(int argc, char const *argv[])
{
int n = 0, a = 0, x = 0, t = 0;
for (int i = 0; i 
{
n = i;
t = 0;
while(n]]></description><category></category></item></channel></rss>