Question

谁可以给我解释一段C语言代码（详细点） 谢谢

/*RSA algorithm */
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MM 7081
#define KK 1789
#define PHIM 6912
#define PP 85
typedef char strtype[10000];
int len;
long nume[10000];
int change[126];
char antichange[37];

void initialize()
{ int i;
char c;
for (i = 11, c = 'A'; c <= 'Z'; c ++, i ++)
{ change[c] = i;
antichange[i] = c;
}
}
void changetonum(strtype str)
{ int l = strlen(str), i;
len = 0;
memset(nume, 0, sizeof(nume));
for (i = 0; i < l; i ++)
{ nume[len] = nume[len] * 100 + change[str[i]];
if (i % 2 == 1) len ++;
}
if (i % 2 != 0) len ++;
}
long binamod(long numb, long k)
{ if (k == 0) return 1;
long curr = binamod (numb, k / 2);
if (k % 2 == 0)
return curr * curr % MM;
else return (curr * curr) % MM * numb % MM;
}
long encode(long numb)
{ return binamod(numb, KK);
}
long decode(long numb)
{ return binamod(numb, PP);
}
main()
{ strtype str;
int i, a1, a2;
long curr;
initialize();
puts("Input 'Y' if encoding, otherwise input 'N':");
gets(str);
if (str[0] == 'Y')
{ gets(str);
changetonum(str);
printf("encoded: ");
for (i = 0; i < len; i ++)
{ if (i) putchar('-');
printf(" %ld ", encode(nume[i]));
}
putchar('\n');
}
else
{ scanf("%d", &len);
for (i = 0; i < len; i ++)
{ scanf("%ld", &curr);
curr = decode(curr);
a1 = curr / 100;
a2 = curr % 100;
printf("decoded: ");
if (a1 != 0) putchar(antichange[a1]);
if (a2 != 0) putchar(antichange[a2]);
}
putchar('\n');
}
putchar('\n');
system("PAUSE");
return 0;
}
对RSA函数的解释

Answer 1

知道MM 7081=73*97，KK 1789，不用多久就能知道PP85了。
RSA要用大素数，这么小的素数手算都能破了你的。

/*RSA algorithm */
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MM 7081
#define KK 1789
#define PHIM 6912
#define PP 85
typedef char strtype[10000];
int len;
long nume[10000];
int change[126];
char antichange[37];

void initialize()
{ int i;
char c;
/*生成字符到数字，和，数字到字符的两张编码表*/
for (i = 11, c = 'A'; c <= 'Z'; c ++, i ++)
{ change[c] = i;
antichange[i] = c;
}
}
/*将字符串转成数字串*/
void changetonum(strtype str)
{ int l = strlen(str), i;
len = 0;
memset(nume, 0, sizeof(nume));
for (i = 0; i < l; i ++)
{ nume[len] = nume[len] * 100 + change[str[i]];
if (i % 2 == 1) len ++; /*每两个字符存在一个int型数中*/
}
if (i % 2 != 0) len ++;
}
/*求（数numb的k次方）mod MM，为了防止溢出，所以每次求次方都进行了模运算*/
/*即去掉函数中的 [% MM] ，剩下的代码就是求数numb的k次方*/
long binamod(long numb, long k)
{ if (k == 0) return 1;
long curr = binamod (numb, k / 2);
if (k % 2 == 0)
return curr * curr % MM;
else return (curr * curr) % MM * numb % MM;
}
/*求（数numb的KK次方）mod MM*/
long encode(long numb)
{ return binamod(numb, KK);
}
/*求（数numb的PP次方）mod MM*/
long decode(long numb)
{ return binamod(numb, PP);
}
main()
{ strtype str;
int i, a1, a2;
long curr;
initialize();
puts("Input 'Y' if encoding, otherwise input 'N':");
gets(str);
if (str[0] == 'Y')
{ gets(str);/*获得字符串*/
changetonum(str); /*字符串转成数字串*/
printf("encoded: ");
for (i = 0; i < len; i ++)
{ if (i) putchar('-');
printf(" %ld ", encode(nume[i]));/*对每个数字加密*/
}
putchar('\n');
}
else
{ scanf("%d", &len);
for (i = 0; i < len; i ++)
{ scanf("%ld", &curr);
curr = decode(curr);/*对每个数字解密*/
a1 = curr / 100; /*数字转成字符*/
a2 = curr % 100;
printf("decoded: ");
if (a1 != 0) putchar(antichange[a1]);
if (a2 != 0) putchar(antichange[a2]);
}
putchar('\n');
}
putchar('\n');
system("PAUSE");
return 0;
}

Answer 2

/*RSA algorithm */ RSA算法 ----- 一个公钥加密的算法

printf("输入要解密的密文数字 c :" );/*输入要解密的密文数字*/
上面这样的解释，真的非常之垃圾，画蛇添足，多此一举！

你还是看看RSA算法的介绍吧，我空间里面有：
http://hi.baidu.com/gaojinshan/blog/item/abf80613cc713b26dd540110.html