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打开源码

分析可知29行在计算flag
29行代码就是利用argv[1]、argv[2]、argv[3]的数据计算

argv[1]

    unsigned int first = atoi(argv[1]);if (first != 0xcafe) {printf("you are wrong, sorry.\n");exit(2);}

如果不等于0xcafe就会退出，所以argv[1]=0xcafe

argv[2]

    unsigned int second = atoi(argv[2]);if (second % 5 == 3 || second % 17 != 8) {printf("ha, you won't get it!\n");exit(3);}

如果满足if条件就会退出，所以要找不满足的第一个数，argv[2]=25

argv[3]

    if (strcmp("h4cky0u", argv[3])) {printf("so close, dude!\n");exit(4);}

如果strcmp里面的数据相等返回0，所以argv[3]=h4cky0u

综上，写出解密代码

#include <stdio.h>
#include <string.h>int main(int argc, char *argv[]) {int first=0xcafe;int second=25; argv[3]="h4cky0u";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;
}