2019 ICPC Malaysia National J.Kitchen Plates

别再乱选离散化方法了!从零到一用Matlab对比差分、双线性Tustin和预矫正的实战效果 本文深入探讨了Matlab中四种主流离散化方法(差分、双线性Tustin和预矫正)的实战效果对比,帮助工程师在数字控制系统设计中做出明智选择。通过详细的理论分析和Matlab代码示例,展示了不同方法在稳定性、频率响应和计算复杂度上的差异,并提供了针对不同场景的工程选用指南。 阅读详情

题目:
https://codeforc.es/gym/102219/problem/J

题意:给ABCDE 5个盘子的5组比较大小,输出完整大小。

这个题应该蛮经典的,题解清一色都是说拓扑排序,可怜我不知道拓扑排序是啥,于是就用了floyd暴力搞的,然后一比较,代码量还差不多。

源代码:

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <climits>
#include <queue>
#include <stack>
#include <map>
//
using namespace std;
const int INF = 0x3f3f3f3f;//1.06e9大小
const int mod1 = 1e9 + 7;
const int mod2 = 998244353;
const int mod3 = 1e9;
const double PI = 3.14159265;
const double eps =1e-8;
typedef unsigned long long ULL;
typedef long long LL;
//
char all[3];
int edge[5][5];
int ans[5];
int main()
{
	for(int time=0;time<5;time++)
	{
		scanf("%s",all);
		if(all[1]=='<')edge[all[2]-'A'][all[0]-'A']=1;
		else edge[all[0]-'A'][all[2]-'A']=1;
	}
	for(int k=0;k<5;++k)
	{
		for(int i=0;i<5;++i)
		{
			for(int j=0;j<5;++j)
			{
				if(edge[i][k]&&edge[k][j])edge[i][j]=1;
			}
		}
	}
	for(int time=0;time<5;time++)
	{
		for(int time1=0;time1<5;time1++)
		{
			if(edge[time][time1]&&edge[time1][time]){printf("impossible\n");return 0;}
		}
	}
	for(int time=0;time<5;time++)
	{
		for(int time1=0;time1<5;time1++)
		{
			if(edge[time][time1])ans[time]++;
		}
	}
	for(int time=0;time<5;time++)
	{
		for(int time1=0;time1<5;time1++)
		{
			if(ans[time1]==time)printf("%c",time1+'A');
		}
	}
	return 0;
}

从上面我用floyd得到点和点之间是否可达,然后用ans[i]表示第i个字母能够直达多少字母。

然后按照顺序输出就行了,超级简洁。

基于遗传算法的柔性作业车间调度(初始化以及编码策略) 这里写自定义目录标题欢迎使用Markdown编辑器新的改变功能快捷键合理的创建标题,有助于目录的生成如何改变文本的样式插入链接与图片如何插入一段漂亮的代码片生成一个适合你的列表创建一个表格设定内容居中、居左、居右SmartyPants创建一个自定义列表如何创建一个注脚注释也是必不可少的KaTeX数学公式新的甘特图功能,丰富你的文章UML 图表FLowchart流程图导出与导入导出导入 欢迎使用Markdown编辑器 你好! 这是你第一次使用 Markdown编辑器 所展示的欢迎页。如果你想学习如何使用Mar 阅读详情

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题目 这个题题意弄明白就会了,也就是对手有一个正整数n,你有一个正整数k<=n,每一次都能对n执行n-=x,(1≤x≤max(1,m−k));把0摆在对方面前就是赢了。 如此我们很容易想起奇偶性博弈,总能把偶数摆在对方面前,我们就行了,所以对所有n,k;我们要做的是,能不能总是把偶数摆在对方面前。 则当k+1== n,或者k== n,每次都只能减去1,所以n为奇数我们才能总是把偶数摆在对方面...

DevourPower的博客 502

NCD 2019题解M. NCD Salary【数学式子处理】

题目 题意,求ab和cd大小关系。均是大数 做一下数学处理:两边取对数,变成blog(a) 和 dlog©;底数是什么无关紧要,为了方便,用c++自带的log10函数就很棒。这里需要特判一下b和d等于0的情况。 int a,b,c,d; double last,now; scanf("%d%d%d%d",&a,&b,&c,&d); if(a==0)las...

DevourPower的博客 472

Codeforces Round #658 (Div. 2) C2 Prefix Flip (Hard Version)【构造匹配】[规定步数内达成目标的题目]

题目 一种操作,可以选择前缀长度为k,翻转前k的01值然后顺序再倒一下。 要求在2n步内把给定a串变成b串 做法: a:01101 01001 b:10000 10001 以这个为例,因为要求是2n步内,很容易就想到每个点搞最多两次搞出来,于是就可以2步之内缩小一次问题规模,从n一直缩小到0; a[1]和b[n]要对上,直接前k翻转. 现在变成谁和谁对上?b[n]和a[n]现在都可以不用管了,于是来到了b[n-1],那b[n-1]和谁对上了,当然还是a[1],a[1]可不就是上一个a[n]吗?可是我们还要用

DevourPower的博客 462

GDUT_寒假训练题解报告_数论专题_个人题解报告——题目:D - Beautiful Numbers (CodeForces - 300C)【逆元、组合数公式】

原题链接http://codeforces.com/problemset/problem/300/C 题目: Vitaly is a very weird man. He’s got two favorite digits a and b. Vitaly calls a positive integer good, if the decimal representation of this int...

DevourPower的博客 444

CF:#620场div2题目题解:B - Longest Palindrome

题目: Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings “pop”, “noon”, “x”, an...

DevourPower的博客 431

Codeforces#642 (Div. 3) D. Constructing the Array

题目链接 题意: 给一个长度为n的空序列,然后往里面填1-n的数字; int cnt=1; 找序列中最长(如果多个最长则取最左边的序列)的连续全空区间:并且赋值cnt++;循环这个操作n次: eg: Consider the array a of length 5 (initially a=[0,0,0,0,0]). Then it changes as follows: Firstly, we choose the segment [1;5] and assign a[3]:=1, so a become

DevourPower的博客 430
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