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：

#include <stdio.h>
#include <stdlib.h>

int prologue [] = {
    0x5920453A, 0x54756F0A, 0x6F6F470A, 0x21643A6F,
    0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
    0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
    0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
    0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
    0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
    0x20206F74, 0x74786565, 0x65617276, 0x32727463,
    0x594E2020, 0x206F776F, 0x79727574, 0x4563200A
};

int data [] = {
    0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
    0x466D203A, 0x65693A72, 0x43646E20, 0x6F54540A,
    0x5920453A, 0x54756F0A, 0x6F6F470A, 0x21643A6F,
    0x594E2020, 0x206F776F, 0x79727574, 0x4563200A,
    0x6F786F68, 0x6E696373, 0x6C206765, 0x796C656B,
    0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
    0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
    0x20206F74, 0x74786565, 0x65617276, 0x32727463,
    0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
    0x21687467, 0x63002065, 0x6C6C7861, 0x78742078,
    0x6578206F, 0x72747878, 0x78636178, 0x00783174
};

int epilogue [] = {
    0x594E2020, 0x206F776F, 0x79727574, 0x4563200A,
    0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
    0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
    0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
    0x20206F74, 0x74786565, 0x65617276, 0x32727463
};

char message[100];

void usage_and_exit(char * program_name) {
    fprintf(stderr, "USAGE: %s key1 key2 key3 key4\n", program_name);
    exit(1);
}

void process_keys12 (int * key1, int * key2) {

    *((int *) (key1 + *key1)) = *key2;
}

void process_keys34 (int * key3, int * key4) {

    *(((int *)&key3) + *key3) += *key4;
}

char * extract_message1(int start, int stride) {
    int i, j, k;
    int done = 0;

    for (i = 0, j = start + 1; ! done; j++) {
        for (k = 1; k < stride; k++, j++, i++) {

            if (*(((char *) data) + j) == '\0') {
                done = 1;
                break;
            }

            message[i] = *(((char *) data) + j);
        }
    }
    message[i] = '\0';
    return message;
}


char * extract_message2(int start, int stride) {
    int i, j;

    for (i = 0, j = start;
         *(((char *) data) + j) != '\0';
         i++, j += stride)
         {
             message[i] = *(((char *) data) + j);
         }
    message[i] = '\0';
    return message;
}

int main (int argc, char *argv[])
{
    int dummy = 1;
    int start, stride;
    int key1, key2, key3, key4;
    char * msg1, * msg2;

    dummy = dummy;
    key3 = key4 = 0;

    if (argc < 3) {
        usage_and_exit(argv[0]);
    }
    key1 = strtol(argv[1], NULL, 0);
    key2 = strtol(argv[2], NULL, 0);
    if (argc > 3) key3 = strtol(argv[3], NULL, 0);
    if (argc > 4) key4 = strtol(argv[4], NULL, 0);

    process_keys12(&key1, &key2);

    start = (int)(*(((char *) &dummy)));
    stride = (int)(*(((char *) &dummy) + 1));

    if (key3 != 0 && key4 != 0) {
        process_keys34(&key3, &key4);
    }

    msg1 = extract_message1(start, stride);

    if (*msg1 == '\0') {
        process_keys34(&key3, &key4);
        msg2 = extract_message2(start, stride);
        printf("%s\n", msg2);
    }
    else {
        printf("%s\n", msg1);
    }

    return 0;
}