A. Night at the Museum
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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 receivedembosser 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:
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
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.
——
题目很水，大意：
指针一开始停到a，然后需要拨到所给的字符串，问需要拨几下走几个格子如zeus 从a走到z需要1，z再走到e需要……（可顺可逆，取最小） 。
最后得需要18次。
（这次英语太渣了，题目背景没看懂，靠看数据猜题意……）
——————
AC代码 ：
#include<cstdio>#include<iostream>#include<cmath>#include<algorithm>using namespace std; int main(){char s ; char now = 'a' ;int ans = 0 ; while ( cin >> s ) {//cout << s << endl ;ans += min ( abs(s - now) , abs( 26 - abs( s - now ) ) );now = s ; } cout << ans << endl ;return 0 ; }
这道题之所以写上来是因为自己实在太蠢了。
一开始一直想不到怎么处理顺时针逆时针怎么取最小？
一开始甚至考虑了 min( abs( s – now ) , abs( now + ‘z’ – ‘a’ +1 – s ) 这么蠢的代码。。
后来感觉这个和约瑟夫环很像，于是考虑用取模，写成了
min( abs ( s – now ) , abs( s – now ) % 26 )
后来发现数据就没有大于26的……怎么取都不出来，然后脑残的还想要不要加一个常数来使这个方法可用。。。
最后突然想到，一般这种情况，对于大于一圈的用mod 小于一圈的用减法就可以了啊！ 被自己蠢哭了。
不过也算一个教训吧，大于一圈以上再考虑mod 小于一圈完全可行。
min( abs( s – now ) , abs( 26 – abs( s – now ) ) ;完破（ 本来第二个表达式用了嵌套的abs是怕 里面答案出现负数，现在感觉我又冗余了……）
