HDoj 2055 An easy problem

Problem Description

we define f(A) = 1, f(a) = -1, f(B) = 2, f(b) = -2, ... f(Z) = 26, f(z) = -26;
Give you a letter x and a number y , you should output the result of y+f(x).

 

 

Input

On the first line, contains a number T.then T lines follow, each line is a case.each case contains a letter and a number.

 

 

Output

for each case, you should the result of y+f(x) on a line.

 

 

Sample Input

6 R 1 P 2 G 3 r 1 p 2 g 3

 

 

Sample Output

19 18 10 -17 -14 -4

 

 

Author

8600

 

 

Source

校庆杯Warm Up

 

 

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注意吸收空格

C语言代码如下：

#include<stdio.h>
int main()
{
    int n=0;
    int y;
    char x;
    scanf("%d",&n);
    getchar();              //注意一定要加getchar();防止后面的scanf(%c)会吸收空格
    int hash[150];
    for(int i=0;i<26;i++)
    {
        hash['A'+i]=i+1;
        hash['a'+i]=-1*(i+1);
    }
    for(int i=0;i<n;i++)
    {
        scanf("%c",&x);       //这句后面不用getchar(),因为%d不会误吸getchar()
        scanf("%d",&y);
        getchar();           //注意这里也一定要加getchar(),防止下一轮循环的scanf("%c")会吸收空格
        printf("%d\n",hash[(int)x]+y);
    }
}