BUUCTF RE section entitled wp

1, easyre
dragged ida, get flag

2，helloworld

Apk file drag processing corrections,
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to give flag

3, xor
dragged ida,
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is a simple XOR, write scripts

glo=[0x66,0x0a,0x6b,0x0c,0x77,0x26,0x4f,0x2e,0x40,0x11,0x78,0x0d,0x5a,0x3b,0x55,0x11,0x70,0x19,0x46,0x1f,0x76,0x22,0x4d,0x23,0x44,0x0e,0x67,0x06,0x68,0x0f,0x47,0x32,0x4f]
x='f'
for i in range(1,len(glo)):
    x+=chr(glo[i]^glo[i-1])
print(x)

Get flag

4, Happy New Year
exe dragged ida, found the shell, check shell
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upx, with upx -d shelling, and then drag ida

get flag

5, reserve1
EXE, dragged ida, there are a number of functions, so look at the string
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you see a suspicious string, the points go

find a function call, and then point to go

find the key function, analysis, string str2 in all o replace 0 is the flag

6,Mysterious
这题出在MISC里了，先找字符串
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跟随

找到关键函数

这里有个atoi()将字符串换为整形之后减一了，所以输入122xyz

得到flag

6,不一样的flag
许久未见的迷宫题，进main
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十分贴心的已经将上下左右分好了，然后看_data_start___这就是迷宫

由*开始，由#结束，每五个一换行

走一遍的flag

7,SimpleRev
拖进ida，可以找到关键函数decry，
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看key1,3，

可以得到 text=killshadow，key=ADSFKNDCLS
然后程序对key进行了一番花里胡哨但其实就是把key变成小写的操作，于是得key=adsfkndcls，而我们要求的是v1，写脚本

#include<stdio.h>
int main(void)
{
        char text[10]="killshadow";
        char key[10]="adsfkndcls";
        int v3=0;
        int v5=10;
        char flag[10];
        for(int i=0;i<10;i++)
        {
                for(char j='A';j<='Z';j++){
                        char temp;
                        temp=(j-39-key[i]+97)%26+97;
                        if(temp==text[i])
                                flag[i]=j;
                }
        }
        printf("%s",flag);
        return 0;
}

Get flag

8, the connotation of software
attendance problems, dragged ida, get flag

BUUCTF RE section entitled wp

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