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攻防世界tt3441810做法(清晰且简单)

news 2026/4/18 9:53:11

简单的就不说了，就一个查壳操作，结果发现既不是32bit也不是64bit，那就随便拖一个进去拖进去之后是这样的里面有一大堆数据，我们选中按A转换为字符串形式，仔细看看，里面有点flag的影子在,那就shift+e提取数据出来，然后放到vs中进行筛选，

#include<stdio.h> int main() { int str[]= { 104, 102, 108, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 97, 103, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 123, 112, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 111, 112, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 112, 111, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 112, 114, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 101, 116, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 125, 10, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 72, 49, 255, 72, 184, 60, 0, 0, 0, 0, 0, 0, 0, 15, 5 }; for (int i = 0; i < sizeof(str) / sizeof(int);i++) { if (str[i] > 32 && str[i] < 128) { printf("%c", str[i]); } } return 0; }

就这样先简单筛选一下看看出来什么东西，注意这边>32&&<128是为了输出字符形式的数据，作为第一次筛选，结果运行出来hflHH4$HHhagHH4$HHh{pHH4$HHhopHH4$HHhpoHH4$HHhprHH4$HHhetHH4$HHh}HH4$HHH1H<

发现有很多的h，H，$，那就筛选掉这些再看看，

#include<stdio.h> int main() { int str[]= { 104, 102, 108, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 97, 103, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 123, 112, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 111, 112, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 112, 111, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 112, 114, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 101, 116, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 104, 125, 10, 0, 0, 72, 191, 1, 0, 0, 0, 0, 0, 0, 0, 72, 141, 52, 36, 72, 186, 2, 0, 0, 0, 0, 0, 0, 0, 72, 184, 1, 0, 0, 0, 0, 0, 0, 0, 15, 5, 72, 49, 255, 72, 184, 60, 0, 0, 0, 0, 0, 0, 0, 15, 5 }; for (int i = 0; i < sizeof(str) / sizeof(int);i++) { if (str[i] > 60 && str[i] < 128 && str[i] != 'h' && str[i] != 'H') { //这边直接取60-128的，去掉特殊字符，只选大小写字母，并去除h，H printf("%c", str[i]); } } return 0; }

然后运行一下flag就直接出来了

总结一下，这道题首先查不出是32，或者是64bit的，这边要敢于操作，直接随便拖一个进去，然后转字符串形式，发现有flag的影子，然后提取数据出去，再运用小算法进行筛选。问题不大，简简单单。

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