Feeding Chicken

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D - Feeding Chicken 从左上角开始,往右下角开始遍历,但是遍历的时候需要注意一点,就是遍历的时候需要连起来,就比如第一行从左往右进行遍历,但是第二行不能从左往右了,因为这样就分开了,所以第二行就应该从右往左遍历,因此,奇数行从左往右遍历,偶数行从右往左遍历. 实现代码: for(int i=1,k=1;i<=r;++i) for(int j=((i&1)?1:c);1<=j&&j<=c;j+=((i&1)?1:-1)) 其他的就很简单了.…

Codeforce 1255 Round #601 (Div. 2)D. Feeding Chicken （模拟）

Long is a huge fan of CFC (Codeforces Fried Chicken). But the price of CFC is increasing, so he decides to breed the chicken on his own farm. His farm is presented by a rectangle grid with rr rows and cc columns. Some of these cells contain rice, oth…

Codeforces Round #601 (Div. 2) D Feeding Chicken

//为了连贯,采取一条路形式,从第一行开始也就是s型 #include <bits/stdc++.h> using namespace std; ; char str[MAXN][MAXN]; vector<char> ch;//存放鸡的名字 void init() { '; i++) ch.emplace_back(i); for(char i='A'; i<='Z'; i++) ch.emplace_back(i); for(char i='a'; i<='z';…

D. Feeding Chicken（构造）

题目大意:将k个鸡放到一个n*m的矩阵中,要求每个鸡所占的rice的个数只差最小题解:构造,设一共有cnt个rice,可以分cnt/k个,即每一只鸡要么占用cnt/k个rice,要么占cnt/k+1个rice.蛇形跑一边矩阵即可. 注意:要判断当前鸡的个数,即如果当前鸡的个数达到k个,那么放置完毕,鸡的数量不能再增长,而且剩下的格子一定是".". #include<bits/stdc++.h> using namespace std; ; char arr[N][N];…

Codeforces Round #601 (Div. 2)

传送门 A. Changing Volume 签到. Code /* * Author: heyuhhh * Created Time: 2019/11/19 22:37:33 */ #include <bits/stdc++.h> #define MP make_pair #define fi first #define se second #define sz(x) (int)(x).size() #define all(x) (x).begin(), (x).end() #define…

读《你必须知道的.NET》继承本质论 Bird bird=new Chicken()

我们创建如下的三层继承层次类. public abstract class Animal { public abstract void ShowType(); } public class Bird : Animal { private string type = "Bird"; public override void ShowType() { Console.WriteLine("Type is {0}", type); } } public class Chi…

2015北京网络赛 A题 The Cats' Feeding Spots 暴力

The Cats' Feeding Spots Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://hihocoder.com/contest/acmicpc2015beijingonline/problem/1 Description In Yan Yuan, the Peking University campus, there are many homeless cats. They all live happy lives because…