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

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

样例输出

191810-17-14-4

 

思路：略。

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