<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[omegaaa的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/omegaaa</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; omegaaa]]></copyright><item><title><![CDATA[[PTA]7-10 计算工资 （15 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85500285</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85500285</guid><author>omegaaa</author><pubDate>Tue, 01 Jan 2019 01:15:00 +0800</pubDate><description><![CDATA[先判断工龄，再判断工时，嵌套条件。
#include&amp;lt;stdio.h&amp;gt;

int main()
{
	int year;
	int time;
	float salary = 0;
	scanf(&quot;%d %d&quot;,&amp;amp;year,&amp;amp;time);
	if( year &amp;lt; 5){
		if(time &amp;gt; 40){
			salary = (time - 40) *45...]]></description><category></category></item><item><title><![CDATA[[PTA]7-9 用天平找小球 （10 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85500154</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85500154</guid><author>omegaaa</author><pubDate>Tue, 01 Jan 2019 01:12:38 +0800</pubDate><description><![CDATA[#include&amp;lt;stdio.h&amp;gt;

int main()
{
	int a, b, c;
	scanf(&quot;%d %d %d&quot;,&amp;amp;a,&amp;amp;b,&amp;amp;c);
	if(a != b &amp;amp;&amp;amp; a != c){
		printf(&quot;A&quot;);
	}else if(a == c){
		printf(&quot;B&quot;);
	}else{
		printf(&quot;C&]]></description><category></category></item><item><title><![CDATA[[PTA]7-8 超速判断 （10 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85500071</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85500071</guid><author>omegaaa</author><pubDate>Tue, 01 Jan 2019 01:11:15 +0800</pubDate><description><![CDATA[同样的简单条件语句，注意输出格式。
#include&amp;lt;stdio.h&amp;gt;

int main()
{
	int speed ;
	scanf(&quot;%d&quot;,&amp;amp;speed);
	if(speed &amp;gt; 60){
		printf(&quot;Speed: %d - Speeding&quot;,speed);
	}else{
		printf(&quot;Speed: %d - OK&quot;,speed);
	}
	...]]></description><category></category></item><item><title><![CDATA[[PTA]7-7 12-24小时制 （15 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85500011</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85500011</guid><author>omegaaa</author><pubDate>Tue, 01 Jan 2019 01:10:05 +0800</pubDate><description><![CDATA[if语句控制三个输出。
#include &amp;lt;stdio.h&amp;gt;

int main()
{
	int hour,minute;
	scanf(&quot;%d:%d&quot;,&amp;amp;hour,&amp;amp;minute);
	if(hour &amp;gt; 12){
		printf(&quot;%d:%d PM&quot;,hour-12,minute);
	}else if(hour == 12){
		printf(&quot;%d:...]]></description><category></category></item><item><title><![CDATA[[PTA]7-6 混合类型数据格式化输入 （5 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85499949</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85499949</guid><author>omegaaa</author><pubDate>Tue, 01 Jan 2019 01:08:52 +0800</pubDate><description><![CDATA[#include &amp;lt;stdio.h&amp;gt;

int main()
{
	float f1,f2;
	int i;
	char c;
	scanf(&quot;%f %d %c %f&quot;,&amp;amp;f1,&amp;amp;i,&amp;amp;c,&amp;amp;f2);
	printf(&quot;%c %d %.2f %.2f&quot;,c,i,f1,f2);
	return 0;
}


]]></description><category></category></item><item><title><![CDATA[[PTA]7-5 表格输出 （5 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85499898</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85499898</guid><author>omegaaa</author><pubDate>Tue, 01 Jan 2019 01:07:41 +0800</pubDate><description><![CDATA[照抄即可。
#include&amp;lt;stdio.h&amp;gt;
#include&amp;lt;math.h&amp;gt;

int main()
{
	printf(&quot;------------------------------------\n&quot;);
	printf(&quot;Province      Area(km2)   Pop.(10K)\n&quot;);
	printf(&quot;-----------------------...]]></description><category></category></item><item><title><![CDATA[[PTA]7-4 BCD解密 （10 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85499851</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85499851</guid><author>omegaaa</author><pubDate>Tue, 01 Jan 2019 01:06:31 +0800</pubDate><description><![CDATA[#include &amp;lt;stdio.h&amp;gt;

int main ()
{
	int n;
	scanf(&quot;%d&quot;,&amp;amp;n);
	printf(&quot;%d&quot;,n / 16 * 10 + n % 16);
	return 0;
}


]]></description><category></category></item><item><title><![CDATA[[PTA]7-3 逆序的三位数 （10 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85499742</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85499742</guid><author>omegaaa</author><pubDate>Tue, 01 Jan 2019 01:04:42 +0800</pubDate><description><![CDATA[既然是三位数，分别将三个位置求出在反向相加即可。
#include&amp;lt;stdio.h&amp;gt;

int main()
{
	int n;
	scanf(&quot;%d&quot;,&amp;amp;n);
	printf(&quot;%d&quot;,(n % 10) * 100+((n % 100) / 10) * 10+(n / 100));
	return 0;
}


...]]></description><category></category></item><item><title><![CDATA[[PTA]7-2 然后是几点 （15 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85499533</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85499533</guid><author>omegaaa</author><pubDate>Tue, 01 Jan 2019 01:00:37 +0800</pubDate><description><![CDATA[将起始、流逝的时间化成分钟，加起来，在分别同60取余、作除法即可。（题目保证起始结束的时间在一天之内）
#include &amp;lt;stdio.h&amp;gt;

int main ()
{
	int ts,tp;
	scanf(&quot;%d %d&quot;,&amp;amp;ts,&amp;amp;tp);
	printf(&quot;%d&quot;,(ts / 100 * 60 + ts % 100 + tp) / 60 * 100 + (ts /...]]></description><category></category></item><item><title><![CDATA[[PTA]7-1 厘米换算英尺英寸 （15 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85499372</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85499372</guid><author>omegaaa</author><pubDate>Tue, 01 Jan 2019 00:56:56 +0800</pubDate><description><![CDATA[比较简单的一道题，照公式求出总英寸数，再分别与12作除、取余即可。此题英寸与英尺均是整形。
#include &amp;amp;lt;stdio.h&amp;amp;gt;

int main()//metre=(foot+inch/12)*0.3048
{
	int centimeter,totalinch;
	scanf(&amp;quot;%d&amp;quot;,&amp;amp;amp;centimeter);
	totalinch = centimeter/30.48...]]></description><category></category></item><item><title><![CDATA[[PTA]7-38 数列求和-加强版 （20 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85464327</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85464327</guid><author>omegaaa</author><pubDate>Mon, 31 Dec 2018 14:34:01 +0800</pubDate><description><![CDATA[基本思路：
1.因为0≤N≤1000000\leq N\leq 1000000≤N≤100000，因此最大数可以达到101010万位，因此考虑用数组解决问题；
2.构造一个新的数组用来存储数列和的每一位，简单观察可知数列求和有如下规律：第iii位是N−iN-iN−i个AAA相加，因此个位是NA10\frac{NA}{10}10NA​的余数，NA10\frac{NA}{10}10NA​的值进位。
3...]]></description><category></category></item><item><title><![CDATA[[PTA]6-12 判断奇偶性 （10 分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85455903</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85455903</guid><author>omegaaa</author><pubDate>Mon, 31 Dec 2018 12:04:36 +0800</pubDate><description><![CDATA[这道题很简单，奇偶判别函数只需要输出1,0。
思路是直接让N除2取余，返回此值的非即可。（奇数除2余1取非为0，偶数除2余0取非为1。）
#include &amp;amp;lt;stdio.h&amp;amp;gt;

int even( int n );

int main()
{    
    int n;

    scanf(&amp;quot;%d&amp;quot;, &amp;amp;amp;n);
    if (even(n))
        printf..]]></description><category></category></item><item><title><![CDATA[[PTA]7-22 龟兔赛跑（20分）]]></title><link>https://blog.youkuaiyun.com/omegaaa/article/details/85395374</link><guid>https://blog.youkuaiyun.com/omegaaa/article/details/85395374</guid><author>omegaaa</author><pubDate>Sun, 30 Dec 2018 16:55:31 +0800</pubDate><description><![CDATA[[PTA]7-22 龟兔赛跑（20分）
C语言的龟兔赛跑问题，用的办法比较笨，逻辑应该还算清晰。
#include &amp;lt;stdio.h&amp;gt;

int main()
{
	int RabbitDistance,T,time=0;//time在这里计数，以10位单位，每10分钟检查一次龟兔的距离。
	scanf(&quot;%d&quot;,&amp;amp;T);
	if(T &amp;lt;= 10){
		printf(&quot;...]]></description><category></category></item></channel></rss>