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java編程:秘聞解碼:原來字母為A,轉換後為D,以此類推, 照片詳情,謝謝哦

#include?<stdio.h>

#defineMAXCHARS1000

//decode?char

char?decode(char?ch);

//decode?string

void?decodes(char?*?s,?int?length);

//fill?zero

void?zero(char?*?s,?int?length);

int?main(void)

{

char?str[MAXCHARS],?ch;

int?count?=?0;

zero(str,?MAXCHARS);

while?((ch?=?getchar())?!=?EOF)?{

str[count++]?=?ch;

if?('\n'?==?ch?||?MAXCHARS?==?count)?{

count?=?0;

decodes(str,?MAXCHARS);

printf("%s\n",?str);

zero(str,?MAXCHARS);

}

}

return?0;

}

char?decode(char?ch)

{

if?(ch?>=?'A'?&&?ch?<=?'W')?{

ch?=?ch?+?3;

}?else?if?(ch?>=?'X'?&&?ch?<=?'Z')?{

ch?=?ch?-?26?+?3;

}?else?{

ch?=?ch;

}

return?ch;

}

void?decodes(char?*?s,?int?length)

{

char?*?t;

for?(int?i=0;?i<length;?i++)?{

*(s?+?i)?=?decode(*(s+i));

}

}

void?zero(char?*?s,?int?length)

{

for?(int?i=0;?i<length;?i++)?{

s[i]?=?0;

}

} Result Like This:

抱歉,看錯,下面是java版:

/**

*?Main.java

*/

import?java.util.Scanner;

public?class?Main?{

public?Main()?{}

public?static?void?main(String[]?args)?{

Scanner?reader?=?new?Scanner(System.in);

while?(reader.hasNext())?{

String?s?=?sc.next();

s?=?Main.decodes(s);

System.out.println(s);

}

}

public?static?char?decode(char?ch)?{

int?ich?=?ch;

if?(ich?>=?'A'?&&?ich?<=?'W')?{

ich?+=?3;

}?else?if?(ich?>=?'X'?&&?ich?<=?'Z')?{

ich?+=?(3-26);

}

return?(char)?ich;

}

public?static?String?decodes(String?str)?{

StringBuffer?sb?=?new?StringBuffer(str);

for?(int?i=0;?i<str.length();?i++)?{

sb.setCharAt(i,?Main.decode(sb.charAt(i)));

}

str?=?sb.toString();

return?str;

}

}

The Result Is:

以上兩個例子都是在Linux下編譯運行,Ctrl+D即可結束運行。

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