算法题:完美平方

给一个正整数 n,写一个函数找到若干个完全平方数(比如 1,4,9,…)使得他们的和等于 n。要求为你需要让平方数的个数最少,输出需要的最少的平方数的个数。

格式:
输入每一行输入一个 整数 n,输出每一行输出需要最少的平方数的个数。
样例输入

n = 12
n = 13

样例输出

3
//12 = 4 + 4 + 4
2
//13 = 4 + 9

解 (用高亮打不出来,日

#include <iostream>
#include <cmath>
using namespace std;
int perfect_square(int);
int main(){
cout<<perfect_square(10)<<endl;
return 0;
}
int perfect_square(int number){
int result = -1;
int start = floor(sqrt(number));
cout<<“start:”<<start<<endl;
if((int)pow(start,2)==number){
return 1;
}
return 1+perfect_square(number-pow(start,2));
}

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js的小技巧(笔记,随时更新)

查看函数源代码:


(函数名).toString();

//复制到剪贴板以便后期处理(Chrome console)

copy((函数名).toString());

萌娘百科:复制歌词原文


var lyric_lines=$('.Lyrics-original');
var lyric_array=[];
lyric_lines.each(function(){
    lyric_array.push(this.innerText);
})

获取humble book bundle 下载地址


var hrefs=[];
$('.download span a').each(function(){hrefs.push(this.href)});
copy(hrefs.join("\r\n"));

MySQL导致关机时间长(Waiting for table flush)

首先说结论:双系统时区同步( timedatectl set-local-rtc 1 )导致

一直以来我这个笔记本在关机的时候都有些慢,本来也没想去管它(因为以前也试图跟开关机的错误作斗争,但我没赢过)

但是随着我的时间越来越紧,也没耐心等它慢慢关机了,就稍微做了个笔记,去搜了一下,大体上是说,mysql在清理垃圾数据,而时区设置导致mysql认为有来自未来(8小时之后)的数据,为了保持数据缓存干净,所以一直在等待,在网络越来越好的情况下,双系统时间同步对我也没那么重要了,所以把这个设置调回了默认值,然后重启系统,好了

后记:这篇记录太早了,才发现没发布出去,大部分都是后来补写的,有没太写清楚的地方,请见谅。