Nine degree OJ 1079 phone keypad (analog)

Topic 1079: Mobile keypad

time limit:1 seconds

Memory limit:32 MB

Special question: No

submitted:1989

Resolution:1069

Title Description:

according to the phone keypad input letters, the calculation of the time spent, such as: A,b,c are in the "1" key, input a only need to press, input C need to press three consecutive times. If the two consecutive characters are not on the same key, you can press directly, such as: ad needs to press two, KZ needs to press 6 if the two consecutive characters on the same key, then two keys between the need to wait a period of time, such as AC, after pressing A, you need to wait a while to press C. It now assumes that a time period is required for each hit, and that the wait time takes two time periods. Now given a string of characters, you need to figure out how much time it takes.

Input:

A string of a length not greater than 100, where only the lowercase letters on the phone keys

Output:

The

input may include multiple sets of data, and for each set of data, the time required for the output to be pressed out by the input string

Sample input:

Bobwww

Sample output:

77

Last hand keyboard diagram

My idea is: to map each letter to the corresponding key, on each key according to the order and distinguish, m[' a ']=10,m[' B ']=11,m[' C ']=12,m[' d ']=20 ...

So the value of the hash table as long as 10 bits is the same, it means that on a key, digit number plus one is to press the number of times

#include <stdio.h> #include <string.h>int ans;char s[120];char m[256];void init () {memset (m,0,sizeof (m));    m[' A ']=m[' B ']=m[' C ']=10;    m[' B ']+=1;    m[' C ']+=2;    m[' d ']=m[' e ']=m[' f ']=20;    m[' E ']+=1;    m[' F ']+=2;    m[' g ']=m[' h ']=m[' I ']=30;    m[' h ']+=1;    m[' i ']+=2;    m[' J ']=m[' K ']=m[' l ']=40;    m[' K ']+=1;    m[' L ']+=2;    m[' m ']=m[' n ']=m[' o ']=50;    m[' n ']+=1;    m[' O ']+=2;    m[' P ']=m[' Q ']=m[' R ']=m[' s ']=60;    m[' Q ']+=1;    m[' R ']+=2;    m[' s ']+=3;    m[' t ']=m[' u ']=m[' V ']=70;    m[' u ']+=1;    m[' V ']+=2;    m[' W ']=m[' x ']=m[' y ']=m[' z ']=80;    m[' x ']+=1;    m[' y ']+=2; M[' z ']+=3;}    int main (int argc, char *argv[]) {//Freopen ("1079.in", "R", stdin);    Init ();        while (~SCANF ("%s", s)) {ans=0;            for (int i=0;i<strlen (s); ++i) {if (i==0) {ans+= (m[s[i]]%10+1);                } else if (M[S[I]]/10==M[S[I-1]]/10) {ans+=2;            ans+= (m[s[i]]%10+1);  } else {              ans+= (m[s[i]]%10+1);    }} printf ("%d\n", ans); } return 0;} /************************************************************** problem:1079 User:kirchhoff language:c Resu lt:accepted time:10 Ms memory:912 kb****************************************************************/

Nine degree OJ 1079 phone keypad (analog)