hdu2055 An easy problem（C语言）

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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

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

Sample Output

191810-17-14-4

Author

8600

Source

校庆杯Warm Up

C语言AC代码

#include<stdio.h>int main(){    int n;    scanf("%d",&n);    getchar();    while(n--)    {        char c;        int y;        scanf("%c %d",&c,&y);        getchar();        if(c>='A'&&c<='Z')            printf("%d\n",y+c-'A'+1);        else            printf("%d\n",y-(c-'a'+1));    }    return 0;}

再上一个牛逼版本的，不是我写的，值得学习！

#include <stdio.h>int main(){    int n, a;    char c;    scanf("%d%*c", &n);    while (n-- && scanf("%c%d%*c", &c, &a))        printf("%d\n", a + (c < 97 ? c - 'A' + 1 : 'a' - c - 1));    return 0;}

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