• a letter and a number


    题目217
    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

     1 #include<stdio.h>
     2 
     3 
     4 
     5  int main(){
     6      int T;
     7       scanf("%d",&T);
     8         while(T--){
     9             int y;
    10              char x;
    11          getchar();  //过滤空格 
    12         scanf("%c %d",&x,&y);
    13         
    14             if(x>='a'&&x<='z')
    15         printf("%d
    ",-(x-96)+y);
    16            if(x>='A'&&x<='Z')
    17        printf("%d
    ",(x-64)+y);
    18          
    19         
    20             
    21         }
    22          
    23          return 0;
    24  }*/
    25 //采用函数调用 
    26  #include<stdio.h>
    27   int f( char x){
    28       if(x>='a'&&x<='z')
    29         return (1);
    30            if(x>='A'&&x<='Z')
    31          return (0);}
    32    int main(){
    33          int T;
    34           scanf("%d",&T);
    35           
    36    while(T--){
    37        int y;
    38              char x;
    39         getchar();
    40         
    41           scanf("%c %d",&x,&y);
    42         if(f(x) )   printf("%d
    ",-(x-96)+y);
    43          if(f(x)==0)      printf("%d
    ",(x-64)+y);
    44   } 
    45          
    46               return 0; 
    47                  } 
    48  
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  • 原文地址:https://www.cnblogs.com/acmgym/p/3691886.html
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