你题目的解答代码如下:
#include <stdio.h>
char fp(char c)
{
if (c >= 'A' && c <= 'Z')
c = (c - 'A' + 4) % 26 + 'A';
if (c >= 'a' && c <= 'z')
c = (c - 'a' + 4) % 26 + 'a';
return c;
}
int main()
{
char c1 = 'T';
char c2 = 'o';
char c3 = 'd';
char c4 = 'a';
char c5 = 'y';
c1 = fp(c1);
c2 = fp(c2);
c3 = fp(c3);
c4 = fp(c4);
c5 = fp(c5);
putchar(c1);
putchar(c2);
putchar(c3);
putchar(c4);
putchar(c5);
putchar('\n');
printf("%c%c%c%c%c\n",c1,c2,c3,c4,c5 );
return 0;
}
如有帮助,望采纳!谢谢!
代码如下:
#include <stdio.h>
//定义这个函数来统一转换
char fun(char c1)
{
if( (c1>='a' && c1 < 'w') || (c1>='A' && c1 < 'W') )
c1 = c1+4;
else if( c1>='w' && c1 <= 'z')
c1 = 'a' + c1+4-'z'-1;
else if( c1>='W' && c1 <= 'Z' )
c1 = 'A' + c1+4-'Z'-1;
return c1;
}
int main()
{
char c1='T',c2='o',c3='d',c4='a',c5='y';
c1 = fun(c1);
c2 = fun(c2);
c3 = fun(c3);
c4 = fun(c4);
c5 = fun(c5);
printf("%c%c%c%c%c",c1,c2,c3,c4,c5);
return 0;
}