codeforces Round #441 A Trip For Meal【思路/模拟】
1 second
512 megabytes
standard input
standard output
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is a meters, between Rabbit's and Eeyore's house is b meters, between Owl's and Eeyore's house is cmeters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.
First line contains an integer n (1 ≤ n ≤ 100) — number of visits.
Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit's and Owl's houses.
Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl's and Eeyore's houses.
Output one number — minimum distance in meters Winnie must go through to have a meal n times.
3
2
3
1
3
1
2
3
5
0
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
【题意】:一个三角形,给出一个数n和三条边长,要求到达三个点总共n次(不要求每个点都到)。起点固定在R,求走的最短距离。
【分析】:在最短路上反复横跳(?)。因为反正可以回头。
【代码】:
#include <bits/stdc++.h>
using namespace std;
int main(void)
{
int n,a,b,c;
scanf("%d%d%d%d",&n,&a,&b,&c);
if(n==)
printf("%d\n", );
if(n==)
printf("%d\n",min(a,b) );
if(n>)
printf("%d\n",min(a,b) + min(min(a,b), c)* (n-) ); return ;
}
codeforces Round #441 A Trip For Meal【思路/模拟】的更多相关文章
- Codeforces Round #441 D. Sorting the Coins(模拟)
http://codeforces.com/contest/876/problem/D 题意:题意真是难懂,就是给一串序列,第i次操作会在p[x](1<=x<=i)这些位置放上硬币,然后从 ...
- Codeforces Round #441 (Div. 2)【A、B、C、D】
Codeforces Round #441 (Div. 2) codeforces 876 A. Trip For Meal(水题) 题意:R.O.E三点互连,给出任意两点间距离,你在R点,每次只能去 ...
- Codeforces Round #441 (Div. 2)
Codeforces Round #441 (Div. 2) A. Trip For Meal 题目描述:给出\(3\)个点,以及任意两个点之间的距离,求从\(1\)个点出发,再走\(n-1\)个点的 ...
- [日常] Codeforces Round #441 Div.2 实况
上次打了一发 Round #440 Div.2 结果被垃圾交互器卡掉 $200$ Rating后心情复杂... 然后立了个 Round #441 要翻上蓝的flag QAQ 晚饭回来就开始搞事情, 大 ...
- Codeforces Round #441 (Div. 2, by Moscow Team Olympiad)
A. Trip For Meal 题目链接:http://codeforces.com/contest/876/problem/A 题目意思:现在三个点1,2,3,1-2的路程是a,1-3的路程是b, ...
- Codeforces Round #441 Div. 2 A B C D
题目链接 A. Trip for Meal 题意 三个点之间两两有路径,分别长为\(a,b,c\),现在从第一个点出发,走\(n-1\)条边,问总路径最小值. 思路 记起始点相邻的边为\(a,b\), ...
- ACM-ICPC (10/16) Codeforces Round #441 (Div. 2, by Moscow Team Olympiad)
A. Trip For Meal Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. ...
- Codeforces Round #284 (Div. 2)A B C 模拟 数学
A. Watching a movie time limit per test 1 second memory limit per test 256 megabytes input standard ...
- Codeforces Round #285 (Div. 2) A B C 模拟 stl 拓扑排序
A. Contest time limit per test 1 second memory limit per test 256 megabytes input standard input out ...
随机推荐
- WebSocket添加事件监听器(6)
WebSocket编程遵循异步编程模型;打开socket后,只需要等待事件发生,而不需要主动向服务器轮询,所以需要在WebSocket对象中添加回调函数来监听事件. WebSocket对象有三个事件: ...
- BZOJ 2820: YY的GCD | 数论
题目: 题解: http://hzwer.com/6142.html #include<cstdio> #include<algorithm> #define N 100000 ...
- THUSC2014酱油记
Day0: 坐飞机到北京,然后报到...跟jason_yu分到一个房间,刚好可以蹭点RP.发现房间460RMB/晚,但再带一份早餐就500RMB,难道早餐是40RMB么...在一家川菜馆吃的午晚餐,感 ...
- 十个迅速提升JQuery性能的技巧
本文提供即刻提升你的脚本性能的十个步骤.不用担心,这并不是什么高深的技巧.人人皆可运用!这些技巧包括: 使用最新版本 合并.最小化脚本 用for替代each 用ID替代class选择器 给选择器指定前 ...
- SCOI2005 互不侵犯 [状压dp]
题目传送门 题目大意:有n*n个格子,你需要放置k个国王使得它们无法互相攻击,每个国王的攻击范围为上下左走,左上右上左下右下,共8个格子,求最多的方法数 看到题目,是不是一下子就想到了玉米田那道题,如 ...
- Spring随笔 —— IOC配置的三种不同方式简介
在spring framework中,IOC的配置是最基础的部分,常见的配置方式有基于xml文件和基于注解的配置方式.除了这两种配置方式之外,今天这里再介绍另一种配置方式,先用小demo重温下我们熟悉 ...
- [05] css优先级
1.优先级计算规则(特殊性) 在css中,有不同的方式编写css,如果想给同一个标签设置样式,选择器的写法有很多种,那么当多个样式都应用于同一个标签,标签优先选择哪个样式呢?按照以下规则: 现有 0, ...
- input 单选按钮radio 取消选中(转载)
input单选按钮: 在radio按钮中添加属性tag 0代表未被选中 HTML代码: <input name="rdo1" value="AA" ty ...
- Nginx各项配置的含义
#user nobody; #配置用户或者组,默认为nobody nobody worker_processes 4; #允许生成的进程数,默认为1 worker_cpu_affinity 00000 ...
- 【数据结构】bzoj1636/bzoj1699排队
Description 每天,农夫 John 的N(1 <= N <= 50,000)头牛总是按同一序列排队. 有一天, John 决定让一些牛们玩一场飞盘比赛. 他准备找一群在对列中为置 ...