2回答

• 

  慕侠8100721

  2017-11-12 16:23:57

  已采纳

  #include <stdio.h>

  int main()

  {

         int x=0,y=0;

          int n;
  

          char a;
  

          printf("请输入要参与运算的字母数，必须2-1000个\n");

          scanf("%d",&x);

           for(int m=0;m<x;m++)

              {

                  scanf("%c   ",&a);

                  if(65<=a<=90)
  

                   n=(int) a-64;
  

                  else if(97<=a<=122)
  

                      n=(int) a-96;
  

                  y=y+n;
  

              }
  

            printf("\n%d",y);

  return 0;

  }  
  

  
  

  7 0

  慕仰8146...

  谢谢了！

  2017-11-13 22:40:31

  共 7 条回复 >

• 

  慕侠8100721

  2017-11-12 17:02:54

  #include <stdio.h>

  int main()

  {

  int x=0,y=0;

  int n;

  char a;

  printf("请输入要参与运算的字母数，必须2-1000个\n");

  scanf("%d",&x);

  for(int m=0;m<x;m++)

  {

  getchar();

  scanf("%c",&a); 

  n= (int) a;

  if(65<=n && n<=90)

  n=n-64;

  else if(97<=n && n<=122)

  n=n-96;

  y=y+n;

  }

    printf("\n%d",y);

  return 0;

  } 

  
  

  0 0