3.《open-source》
根据题目指点,编译后直接对源代码进行分析,下面的注释就是解题过程
#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[]) {
if (argc != 4) { //判断输入参数,这里除程序名还需要写3个
printf("what?\n");
exit(1);
}
unsigned int first = atoi(argv[1]); //第一处输入十进制51966
if (first != 0xcafe) {
printf("you are wrong, sorry.\n");
exit(2);
}
unsigned int second = atoi(argv[2]);
if (second % 5 == 3 || second % 17 != 8) { //第二处逻辑判断,前后两个条件都不能符合
printf("ha, you won't get it!\n");
exit(3);
}
if (strcmp("h4cky0u", argv[3])) { //第三处比较判断,直接输入h4cky0u
printf("so close, dude!\n");
exit(4);
}
printf("Brr wrrr grr\n");
unsigned int hash = first * 31337 + (second % 17) * 11 + strlen(argv[3]) - 1615810207;
printf("Get your key: ");
printf("%x\n", hash);
return 0;
}
第二处的脚本
for a range (0,100):
if (a % 5 != 3 and a % 17 == 8):
print (a)
得出a可为25,然后得到flag
4.《insanity》
我醉了。。。小朋友你是否有很多问号