nyo127

最新推荐文章于 2018-09-07 16:48:39 发布

shengweisong

最新推荐文章于 2018-09-07 16:48:39 发布

阅读量611

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/shengweisong/article/details/27070823

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

     
      查看代码---运行号：716626----结果：Accepted
      
       运行时间：
       2014-02-06 21:53:49  |  运行人:
       shengweisong
      
     
     
     
      
       
        
         view source
         
          
         
         print
         ?
        
       
       
        
         01.
         #include<stdio.h>
        
        
         02.
         //#include<stdlib.h> 去掉// 时间会增加
        
        
         03.
         int main()
        
        
         04.
         {
        
        
         05.
         int x, y;
        
        
         06.
         char a;
        
        
         07.
         scanf( "%d", &x );
        
        
         08.
         while( x -- ){
        
        
         09.
         getchar();
        
        
         10.
         scanf( "%c%d", &a, &y );
        
        
         11.
         if( a >='a'&& a<='z')
        
        
         12.
         printf( "%d\n", y -( a -'a'+1 ) );
        
        
         13.
         else
        
        
         14.
         printf( "%d\n", y + a- 'A'+1 );
        
        
         15.
         }
        
        
         16.
         // system ( "pause" );
        
        
         17.
         return 0;
        
        
         18.
         }