nyoj 217-a letter and a number (char)

217-a letter and a number

内存限制:64MB
时间限制:3000ms
特判: No

通过数:4
提交数:5
难度:1

题目描述:

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).

输入描述:

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).

输出描述:

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

样例输入:

复制

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

样例输出:

19
18
10
-17
-14
-4

C/C++ AC：

 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <stack>
 #include <set>
 #include <map>
 #include <queue>
 #include <climits>
 #define PI 3.1415926

 using namespace std;
 int n;

 int main()
 {
     cin >>n;
     while (n --)
     {
         char letter;
         int number;
         cin >>letter >>number;
         if (letter >= 'A' && letter <= 'Z')
         {
             cout <<letter - 'A' +  + number <<endl;
         }
         else
         {
             cout <<- * (letter - 'a' + ) + number <<endl;
         }
     }
 }