CodeForces 731A - Night at the Museum(模拟)

本文介绍了一个有趣的编程挑战,任务是在限定的步数内通过旋转字母轮盘来拼写出展品的名字,并提供了一段高效的实现代码。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

A. Night at the Museum
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.

Embosser is a special devise that allows to “print” the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it’s allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter ‘a’. Other letters are located as shown on the picture:

这里写图片描述

After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It’s not required to return the wheel to its initial position with pointer on the letter ‘a’.

Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.

Input
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It’s guaranteed that the string consists of only lowercase English letters.

Output
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.

Examples
input
zeus
output
18
input
map
output
35
input
ares
output
34
Note

这里写图片描述

To print the string from the first sample it would be optimal to perform the following sequence of rotations:

from ‘a’ to ‘z’ (1 rotation counterclockwise),
from ‘z’ to ‘e’ (5 clockwise rotations),
from ‘e’ to ‘u’ (10 rotations counterclockwise),
from ‘u’ to ‘s’ (2 counterclockwise rotations).
In total, 1 + 5 + 10 + 2 = 18 rotations are required.

题意:
给出一个字符串,求出每个字符里上个字符的最近距离(26个字母围成一个环)的和。

解题思路:
模拟。

AC代码:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    char s[105];
    gets(s);
    int res = 0;
    char tmp = 'a';
    for(int i = 0;s[i];i++) res += min(max(tmp,s[i])-min(tmp,s[i]),min(tmp,s[i])+26-max(tmp,s[i])),tmp = s[i];
    printf("%d\n",res);
    return 0;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值