<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[mzpppp的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/mzpppp</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; mzpppp]]></copyright><item><title><![CDATA[高精度计算]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/159694898</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/159694898</guid><author>mzpppp</author><pubDate>Tue, 31 Mar 2026 20:32:15 +0800</pubDate><description><![CDATA[加法：for(i=0;i<m;i++)t=t+a[i];if(i<n)t=t+b[i];c[i]=t%10;t=t/10;if(t)c[i]=1;j>=0;j--)elsej>=0;j--)int main()i>=0;i--)i>=0;i--)减法：return 1;i++)if(a[i]!=b[i])return 1;]]></description><category></category></item><item><title><![CDATA[经典位运算和计算各进制下的各位数字之和]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/159475567</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/159475567</guid><author>mzpppp</author><pubDate>Wed, 25 Mar 2026 21:22:20 +0800</pubDate><description><![CDATA[(num & (num - 1)) 是检测2的幂的经典位运算方法，结果为0即为2的幂 if ((num & (num - 1))!]]></description><category></category></item><item><title><![CDATA[和数的各位数字有关的题目]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/159463828</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/159463828</guid><author>mzpppp</author><pubDate>Wed, 25 Mar 2026 11:24:32 +0800</pubDate><description><![CDATA[int ans=0;i++)int num=i;return 0;int ans=0;i<=202504;i++)int num=i;int tmp=0;num/=10;ans++;return 0;]]></description><category></category></item><item><title><![CDATA[一维坐标下的最小总距离分配]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/159434610</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/159434610</guid><author>mzpppp</author><pubDate>Tue, 24 Mar 2026 20:48:43 +0800</pubDate><description><![CDATA[int main()int n=0;i<n;i++)i<n;i++)ll ans=0;i<n;i++)return 0;]]></description><category></category></item><item><title><![CDATA[暴力判断是否为质数]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/159252869</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/159252869</guid><author>mzpppp</author><pubDate>Thu, 19 Mar 2026 21:40:00 +0800</pubDate><description><![CDATA[x >= i * i;i++)return 0;return 1;]]></description><category></category></item><item><title><![CDATA[【无标题】]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/159172408</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/159172408</guid><author>mzpppp</author><pubDate>Tue, 17 Mar 2026 19:43:30 +0800</pubDate><description><![CDATA[/ 关键修正：读取n和m的值 scanf("%d %d", &n, &m);]]></description><category></category></item><item><title><![CDATA[空间切割问题]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/155323031</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/155323031</guid><author>mzpppp</author><pubDate>Thu, 27 Nov 2025 21:07:14 +0800</pubDate><description><![CDATA[把上面所有等式的「左边相加」和「右边相加」，得到：g(1)+g(2)+g(3)+...+g(k−1)​=[g(0)+1]+[g(1)+2]+[g(2)+3]+...+[g(k−2)+(k−1)]​。递推公式：g(k) = g(k-1) + k（初始条件：g(0)=1，0 条直线时平面是 1 个区域）。递推公式：p(k) = p(k-1) + 4(k-1) + 1（初始条件：p(0)=1，p(1)=2)通项公式：p(k) = 2k² - k + 1。时：g(k−1)=g(k−2)+(k−1)]]></description><category></category></item><item><title><![CDATA[并查集和最小生成树]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/154848594</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/154848594</guid><author>mzpppp</author><pubDate>Fri, 14 Nov 2025 21:54:30 +0800</pubDate><description><![CDATA[/ 若根不同，将a的根挂到b的根下（简单合并，不考虑树的高度）if (used == n-1) printf("%lld\n", ans);= findp(v)) {  // 若u和v不在同一集合（不连通）void unite(int a,int b){  // 合并a和b所在的集合（简化版，无按秩合并）]]></description><category></category></item><item><title><![CDATA[三分查找问题]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/154202863</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/154202863</guid><author>mzpppp</author><pubDate>Fri, 31 Oct 2025 20:45:43 +0800</pubDate><description><![CDATA[函数需要满足：在定义域内是单峰函数，先严格递增后严格递减，存在唯一最大值，或先严格递减后严格递增，存在唯一最小值；且在定义域内连续且可比较大小。若 f(m1) < f(m2)：说明最大值在 [m1, right] 内（左侧区间 [left, m1] 可舍弃）若 f(m1) > f(m2)：说明最大值在 [left, m2] 内（右侧区间 [m2, right] 可舍弃）若 f(m1) == f(m2)：最大值在 [m1, m2] 内（两端区间可舍弃），函数在该区间内先增后减。这时候可以使用三分查找。]]></description><category></category></item><item><title><![CDATA[二分查找问题]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/154202226</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/154202226</guid><author>mzpppp</author><pubDate>Fri, 31 Oct 2025 20:18:18 +0800</pubDate><description><![CDATA[使用条件：待查找的序列必须是有序的，且单调递增或者单调递减,用于有序序列或者已经排序的数组。目标值，说明目标在左半区间，更新右边界（目标值，说明目标在右半区间，更新左边界（重复上述步骤，直到找到目标或区间无效（习惯于用中间值mid来对比。（数组下标从 0 开始），目标值，查找成功，返回。，恰好等于目标，返回。]]></description><category></category></item><item><title><![CDATA[最大公约数和最小公倍数]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/154201664</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/154201664</guid><author>mzpppp</author><pubDate>Fri, 31 Oct 2025 19:48:07 +0800</pubDate><description><![CDATA[要掌握LCM（最小公倍数） 与 GCD（最大公约数） 的关系（LCM (a,b) = a*b / GCD (a,b)）]]></description><category></category></item><item><title><![CDATA[找规律运算]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/154201565</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/154201565</guid><author>mzpppp</author><pubDate>Fri, 31 Oct 2025 19:34:39 +0800</pubDate><description><![CDATA[通常先带入几个数字进去，一般从1开始带入，算出来结果，寻找规律，用函数来展示出来。]]></description><category></category></item><item><title><![CDATA[快速幂运算内容]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/154201478</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/154201478</guid><author>mzpppp</author><pubDate>Fri, 31 Oct 2025 19:28:56 +0800</pubDate><description><![CDATA[/a %= mod;while (b)b = b / 2;return ans;int main()int n;return 0;while (b)b = b / 2;return ans;int main()int n;= EOF)int a = 1;= 1)a++;return 0;]]></description><category></category></item><item><title><![CDATA[10.26上午leetcode周赛]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/153980685</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/153980685</guid><author>mzpppp</author><pubDate>Mon, 27 Oct 2025 21:01:22 +0800</pubDate><description><![CDATA[include <stdlib.h>    包含标准库函数，如字符串转整数的函数astoll。idx是数组res下标的索引值，res[idx++]=str[i]相当于res[idx]=str[i];sprintf作用是：格式化输出函数，将n按照%lld（长整型格式）转换为字符串，结果存储在str字符数组中。使用long long长整型，可以输入更长的数字，同时返回的时候也是long long类型，防止溢出。思路：构造一个新函数，输入一组数字，使其在遇到非0的数字的时候输出。]]></description><category></category></item><item><title><![CDATA[25.10.22]]></title><link>https://blog.youkuaiyun.com/mzpppp/article/details/153774608</link><guid>https://blog.youkuaiyun.com/mzpppp/article/details/153774608</guid><author>mzpppp</author><pubDate>Mon, 27 Oct 2025 19:54:58 +0800</pubDate><description><![CDATA[使用最小值的思想，1.当牛可乐出石头的时候，只需要看牛牛出了多少次的布（因为题目要求的是牛牛赢的次数，故出剪刀和石头的次数都不计入其中）取两者的较小值；2.可乐出剪刀只用看牛牛出的石头次数；3.可乐出布只用看牛牛出剪刀的次数，依旧取两者之间的较小值。616是由2个6和1个1，我们需要统计数字串中的6的数量cnt6和1的数量cnt1。= '\0' 理论上相当于i<n，  C 语言没有专门的 “字符串类型”，字符串本质是字符数组，但必须在数组的最后一个有效字符后面，额外存储一个特殊字符。]]></description><category></category></item></channel></rss>