zoukankan      html  css  js  c++  java
  • 每日算法

    每日算法

    those times when you get up early and you work hard; those times when you stay up late and you work hard; those times when don’t feel like working — you’re too tired, you don’t want to push yourself — but you do it anyway. That is actually the dream. That’s the dream. It’s not the destination, it’s the journey. And if you guys can understand that, what you’ll see happen is that you won’t accomplish your dreams, your dreams won’t come true, something greater will. mamba out


    那些你早出晚归付出的刻苦努力,你不想训练,当你觉的太累了但还是要咬牙坚持的时候,那就是在追逐梦想,不要在意终点有什么,要享受路途的过程,或许你不能成就梦想,但一定会有更伟大的事情随之而来。 mamba out~

    2020.3.29


    我太菜了 今天开始补cfdiv3

    CF#629Div3 A

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    
    using namespace std;
    int n;
    int main()
    {
    	cin >> n;
    	int a, b ;
    	while(n --)	
    	{
    		cin >> a >> b;
    		if(a % b == 0)printf("0
    ");
    		else{ 
    			printf("%d
    ",b - (a % b));
    		}
    	}
    	return 0;
    } 
    

    CF#629Div3 B

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    #include<cmath>
     
    using namespace std;
     
    int main()
    {
    	int t, ii, n, k, sum, i, j;
    	scanf("%d", &t);
    	for (ii = 0; ii<t; ii++) 
    	{
    		scanf("%d %d", &n, &k);
    		sum = 0;
    		k--;
    		for (i = 0;; i++)
    		{
    			if (sum + i + 1 > k)
    			{
    				break;
    			}
    			sum += i + 1;
    		}
    		k -= sum;
    		i++;
    		for (j = 0; j<n; j++) 
    		{
    			if (j == n - i - 1 || j == n - k - 1)
    			{
    				printf("b");
    			}
    			else
    			{
    				printf("a");
    			}
    		}
    		printf("
    ");
    	}
    	return 0;
    }
    

    CF#629Div3 C

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    
    using namespace std;
    
    void func()
    {
    	int n;
    	cin >> n;
    	string s , a , b;
    	cin >> s;
    	int flag = 0;
    	for(int i = 0;i < s.size() ;i ++)
    	{
    		if(s[i] == '2')
    		{
    			if(flag == 0)
    				a+='1',b+='1';
    			else 
    				a+='0',b+='2';	
    		}
    		else if(s[i] == '0')a+='0',b+='0';
    		else{
    			if(flag == 0)
    			{
    				flag = 1;
    				a+='1';b+='0'; 
    			}
    			else{
    				a += '0';b += '1';
    			}
    		}	
    	}
    	cout << a << endl;
    	cout << b << endl;	
    }
    int main()
    {
    	int t;cin >> t;
    	while(t--)
    	{
    		func();	
    	}	
    	return 0;
    } 
    

    摆动序列

    假dp 真搜索

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    
    using namespace std;
    int n, ans;
    bool vis[25];
    void dfs(int a ,int b)
    {
    	if(b > a){
    		for(int i = 1;i < a ;i ++)
    		{
    			if(!vis[i])
    			{
    				ans++;
    				vis[i] = 1;
    				dfs(b , i);
    				vis[i] = 0;
    			}
    		}
    	}
    	else if(a > b){ 
    		for(int i = a + 1;i <= n;i ++)
    		{
    			if(!vis[i])
    			{	
    				ans++;
    				vis[i] = 1;
    				dfs(b , i);
    				vis[i] = 0;
    			}
    		}
    	}
    }
    int main()
    {
    	cin >> n;
    	for(int i = 1;i <= n ;i ++)
    	{
    		vis[i] = 1;
    		for(int j = 1;j <= n ;j ++)
    		{
    			if(i != j)
    			{
    				ans++;
    				vis[j] = 1;
    				dfs(i , j);
    				vis[j] = 0;
    			}
    		}
    	}
    	cout << ans << endl;
    	return 0;
    }
    
    

    拦截导弹

    (o(n^{2}))

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    
    using namespace std;
    const int N = 100005;
    int f[N], a[N], n;
    
    int main()
    {
    	while(~scanf("%d",&a[++n]));n--;
    	int ans1 = 0, ans2 = 0;
    	for(int i = n;i >= 1 ;i --)
    	{
    		f[i] = 1;
    		for(int j = i + 1;j <= n ;j ++)
    		{
    			if(a[j] <= a[i])
    				f[i] = max(f[i],f[j] + 1);
    		}
    		ans1 = max(ans1,f[i]);
    	}
    	for(int i = 1;i <= n ;i ++)
    	{
    		f[i] = 1;
    		for(int j = 1;j < i ;j ++)
    		{
    			if(a[j] < a[i])
    				f[i] = max(f[i],f[j] + 1);
    		}
    		ans2 = max(ans2,f[i]);
    	}
    	printf("%d
    %d",ans1,ans2);
    	return 0;
    }
    
  • 相关阅读:
    树形DP求树的最小支配集,最小点覆盖,最大独立集
    贪心法求树的最小支配集,最小点覆盖,最大独立集
    树上两点的最近公共祖先问题(Least Common Ancestors)
    大厂前端面试题
    表单中包含上传图片
    element-ui表单验证
    一、React基础
    点击div上传图片,在img中预览
    持久化
    docker安装
  • 原文地址:https://www.cnblogs.com/wlw-x/p/12595469.html
Copyright © 2011-2022 走看看