题目名字叫做开源,然而二者之间并没有什么关系…
一个很简答的C程序题,让你跳过所有的限制条件即可输出key(flag)。
open-source如下:
#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[]) {
if (argc != 4) {
printf("what?\n");
exit(1);
}
unsigned int first = atoi(argv[1]);
if (first != 0xcafe) {
printf("you are wrong, sorry.\n");
exit(2);
}
unsigned int second = atoi(argv[2]);
if (second % 5 == 3 || second % 17 != 8) {
printf("ha, you won't get it!\n");
exit(3);//25
}
if (strcmp("h4cky0u", argv[3])) {
printf("so close, dude!\n");
exit(4);
}
printf("Brr wrrr grr\n");
unsigned int hash = first * 31337 + (second % 17) * 11 + strlen(argv[3]) - 1615810207;
//first=0xcafe,(second%17)=8,argv[3]="h4cky0u"
printf("Get your key: ");
printf("%x\n", hash);
return 0;
}
在原程序上进行一定的修改,得到:
int main()
{
unsigned int hash = 0xcafe * 31337 + 8 * 11 + 7 - 1615810207;
printf("Get your key: ");
printf("%x\n", hash);
return 0;
}
flag:c0ffee