攻防世界逆向新手题3道

1.open-soure

这个直接给了源代码

#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);
    }

    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;

    printf("Get your key: ");
    printf("%x\n", hash);
    return 0;
}

unsigned int hash = first * 31337 + (second % 17) * 11 + strlen(argv[3]) - 1615810207;

根据这行代码，得出要求出first、second、arvg[3]

 if (first != 0xcafe) {
    	printf("you are wrong, sorry.\n");
    	exit(2);
    }

上述代码得出，first=0xcafe，是个16进制的数
转化十进制=51966

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

上述代码得出second=25

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

上述代码得出argv[3]与h4cky0u相等，说明argv[3]=h4cky0u

#include <stdio.h>
#include <string.h>

int main(){
unsigned int hash =  51966* 31337 + (25% 17) * 11 + strlen("h4cky0u") - 1615810207;
printf("Get your key: ");
    printf("%x\n", hash);
    return 0;
}

写个代码运行一下得出

flag就是{c0ffee}

2.simple-unpack

这个是要脱壳，首先判断什么壳

发现是upx壳

用upx脱壳成功，放进ida中查看，找的main函数，直接就出来了

3.insanity

这个就是下载附件放入32位ida中
F5查看字符串