【Codeforces Round #376 (Div. 2)】 Codeforces 731A Night at the Museum

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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:

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.

模拟。

#include<cstdio>
#include<cstring>
char s[110];
int abs(int x)
{return x>0?x:-x;
}
int min(int x,int y)
{return x<y?x:y;
}
int main()
{int i,j,k,x,y,z,n,ans=0,l,p;scanf("%s",s+1);l=strlen(s+1);p='a';for (i=1;i<=l;i++){ans+=min(abs(s[i]-p),26-abs(s[i]-p));p=s[i];}printf("%d\n",ans);
}