/*实践再次说明ch=getchar()的速度非常慢*/

/*大水题，不解释*/

#include<stdio.h>

#include<string.h>

int main()

{

    int i,j,n,count;

    int a[];

    char ch,s[];

    j=;

    for(i=; i<=; i++)

    {

        if(j>) j=;

        a[i]=j;

        j++;

    }

    a[]=;

    a[]=;

    a[]=;

    a[]=;

    a[]=;

    a[]=;

    a[]=;

    a[]=;

    a[]=;

    scanf("%d",&n);

    getchar();

    for(i=; i<=n; i++)

    {

        count=;

       gets(s);

       for(j=;j<strlen(s);j++)

       count+=a[s[j]];

        printf("%d\n",count);

    }

    return ;

}

时间限制:500MS 内存限制:1000K 提交次数:95 通过次数:27

题型: 编程题语言: C++;C

Description

Now almost every student has a mobile telephone. But do you have making attention to the keyboard of a mobile telephone? This is the keyboard of a normal mobile telephone：

It will show different letters if you press one key different times, for example you press the key “2” once it will show a letter ‘a’, twice show a ‘b’. And notice that press the key ‘0’ once will show a space. Now give you a sentence, which contains only lowercases and space, calculate how much times to press the keyboard at least to show the sentence.

输入格式

The first line contains an integer n, which means the number of cases. Per case consist of only one sentence, which contains only lowercases and spaces and its length less or equal 200.

输出格式

Per case output an integer, which is the least times to press the keyboard, in one line.

输入样例

1this problem is so easy

输出样例

Time:15ms

巴特西

6754 Keyboard of a Mobile Telephone

Description

输入格式

输出格式

输入样例

输出样例

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