nyoj 217 a letter and a number

a letter and a number

时间限制：3000 ms | 内存限制：65535 KB
难度：1

描述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).

输入On the first line, contains a number T(0<T<=10000).then T lines follow, each line is a case.each case contains a letter x and a number y(0<=y<1000).
输出for each case, you should the result of y+f(x) on a line
样例输入

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

样例输出

19
18
10
-17
-14
-4

#include<stdio.h>

int main(int argc,char * argv[]){

int N = 0;

scanf("%d",&N);

N=2*N;

while(N--){

char arr;

int n = 0;

scanf("%c %d",&arr,&n);

if(arr >= 'A'&& arr <='Z')

printf("%d\n",n +arr - 'A'+1);

else if(arr >= 'a'&& arr <= 'z')

printf("%d\n",n-(arr -'a')-1);

}

return 0 ;

}