一道老外的破题

这个题目是要求运行时输入4个参数,就是四个key,先把key1,key2输对,会打出一段话,提示下一步的工作要点,把key3,key4输对了,会看到成功的提示,key1,key4编译器相关,说出key2,key3就行了.源码就在下面了,大家来试试吧.
 代码如下：

1. #include <stdio.h>
   
2. #include <stdlib.h>
   
3. 
   
4. int prologue [] = {
   
5.     0x5920453A, 0x54756F0A, 0x6F6F470A, 0x21643A6F,
   
6.     0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
   
7.     0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
   
8.     0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
   
9.     0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
   
10.     0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
    
11.     0x20206F74, 0x74786565, 0x65617276, 0x32727463,
    
12.     0x594E2020, 0x206F776F, 0x79727574, 0x4563200A
    
13. };
    
14. 
    
15. int data [] = {
    
16.     0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
    
17.     0x466D203A, 0x65693A72, 0x43646E20, 0x6F54540A,
    
18.     0x5920453A, 0x54756F0A, 0x6F6F470A, 0x21643A6F,
    
19.     0x594E2020, 0x206F776F, 0x79727574, 0x4563200A,
    
20.     0x6F786F68, 0x6E696373, 0x6C206765, 0x796C656B,
    
21.     0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
    
22.     0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
    
23.     0x20206F74, 0x74786565, 0x65617276, 0x32727463,
    
24.     0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
    
25.     0x21687467, 0x63002065, 0x6C6C7861, 0x78742078,
    
26.     0x6578206F, 0x72747878, 0x78636178, 0x00783174
    
27. };
    
28. 
    
29. int epilogue [] = {
    
30.     0x594E2020, 0x206F776F, 0x79727574, 0x4563200A,
    
31.     0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
    
32.     0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
    
33.     0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
    
34.     0x20206F74, 0x74786565, 0x65617276, 0x32727463
    
35. };
    
36. 
    
37. char message[100];
    
38. 
    
39. void usage_and_exit(char * program_name) {
    
40.     fprintf(stderr, "USAGE: %s key1 key2 key3 key4\n", program_name);
    
41.     exit(1);
    
42. }
    
43. 
    
44. void process_keys12 (int * key1, int * key2) {
    
45.      
    
46.     *((int *) (key1 + *key1)) = *key2;
    
47. }
    
48. 
    
49. void process_keys34 (int * key3, int * key4) {
    
50. 
    
51.     *(((int *)&key3) + *key3) += *key4;
    
52. }
    
53. 
    
54. char * extract_message1(int start, int stride) {
    
55.     int i, j, k;
    
56.     int done = 0;
    
57. 
    
58.     for (i = 0, j = start + 1; ! done; j++) {
    
59.         for (k = 1; k < stride; k++, j++, i++) {
    
60. 
    
61.             if (*(((char *) data) + j) == '\0') {
    
62.                 done = 1;
    
63.                 break;
    
64.             }
    
65.                               
    
66.             message[i] = *(((char *) data) + j);
    
67.         }
    
68.     }
    
69.     message[i] = '\0';
    
70.     return message;
    
71. }
    
72. 
    
73. 
    
74. char * extract_message2(int start, int stride) {
    
75.     int i, j;
    
76. 
    
77.     for (i = 0, j = start;
    
78.          *(((char *) data) + j) != '\0';
    
79.          i++, j += stride)
    
80.          {
    
81.              message[i] = *(((char *) data) + j);
    
82.          }
    
83.     message[i] = '\0';
    
84.     return message;
    
85. }
    
86. 
    
87. int main (int argc, char *argv[])
    
88. {
    
89.     int dummy = 1;
    
90.     int start, stride;
    
91.     int key1, key2, key3, key4;
    
92.     char * msg1, * msg2;
    
93. 
    
94.     dummy = dummy;
    
95.     key3 = key4 = 0;
    
96. 
    
97.     if (argc < 3) {
    
98.         usage_and_exit(argv[0]);
    
99.     }
    
100.     key1 = strtol(argv[1], NULL, 0);
     
101.     key2 = strtol(argv[2], NULL, 0);
     
102.     if (argc > 3) key3 = strtol(argv[3], NULL, 0);
     
103.     if (argc > 4) key4 = strtol(argv[4], NULL, 0);
     
104. 
     
105.     process_keys12(&key1, &key2);
     
106. 
     
107.     start = (int)(*(((char *) &dummy)));
     
108.     stride = (int)(*(((char *) &dummy) + 1));
     
109. 
     
110.     if (key3 != 0 && key4 != 0) {
     
111.         process_keys34(&key3, &key4);
     
112.     }
     
113. 
     
114.     msg1 = extract_message1(start, stride);
     
115. 
     
116.     if (*msg1 == '\0') {
     
117.         process_keys34(&key3, &key4);
     
118.         msg2 = extract_message2(start, stride);
     
119.         printf("%s\n", msg2);
     
120.     }
     
121.     else {
     
122.         printf("%s\n", msg1);
     
123.     }
     
124. 
     
125.     return 0;
     
126. }