Codeforces Round #376 (Div. 2) A. Night at the Museum【模拟】
1 second
256 megabytes
standard input
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.
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.
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
zeus
18
map
35
ares
34
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.
原题链接：http://codeforces.com/contest/731/problem/A
题意：一个圆形的字母盘，一开始指针位于'a'的位置，问你转出该序列上的字母至少需要转动多少次。
由于是环形的，可以顺时针转动或逆时针，每次转动前取最小的就可以了。
AC代码：
/**
* 行有余力，则来刷题！
* 博客链接：http://blog.csdn.net/hurmishine
*
*/
#include <bits/stdc++.h>
using namespace std;
int main()
{
//freopen("data.txt","r",stdin);
char a[105];
while(cin>>a)
{
int len=strlen(a);
char p='a';
int ans=0;
for(int i=0;a[i];i++)
{
int t=abs(a[i]-p);
//cout<<min(t,26-t)<<endl;
ans+=min(t,26-t);
p=a[i];
}
cout<<ans<<endl;
}
return 0;
}
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