【buuctf】findit_buuctf findit-优快云博客

本文链接：https://blog.youkuaiyun.com/ywx05_/article/details/145873182

下载后看到是apk

用jadx打开

找到main函数看看

分析代码应该是对字符串进行了凯撒加密

将字符串提取出来

pvkq{m164675262033l4m49lnp7p9mnk28k75}
f到p=10，l到v=10，那么得出凯撒的位移为10

借助网站帮我们破解

也可以写个解密脚本

#include<iostream>
using namespace std;
int main()
{
	char a[]={'T', 'h', 'i', 's', 'I', 's', 'T', 'h', 'e', 'F', 'l', 'a', 'g', 'H', 'o', 'm', 'e'};
	char b[]= {'p', 'v', 'k', 'q', '{', 'm', '1', '6', '4', '6', '7', '5', '2', '6', '2', '0', '3', '3', 'l', '4', 'm', '4', '9', 'l', 'n', 'p', '7', 'p', '9', 'm', 'n', 'k', '2', '8', 'k', '7', '5', '}'};
	int x[18]={0};
	int y[39]={0};
	for (int i = 0; i < 17; i++) {
		if ((a[i] < 'I' && a[i] >= 'A') || (a[i] < 'i' && a[i] >= 'a')) {
			x[i] = (char) (a[i] + 18);
		} else if ((a[i] < 'A' || a[i] > 'Z') && (a[i] < 'a' || a[i] > 'z')) {
			x[i] = a[i];
		} else {
			x[i] = (char) (a[i] - '\b');
		}
	}
	for (int i2 = 0; i2 < 38; i2++) 
	{
		if ((b[i2] < 'A' || b[i2] > 'Z') && (b[i2] < 'a' || b[i2] > 'z'))
		{
			y[i2] = b[i2];
		}
		else {
			y[i2] = b[i2] + 16;
			if ((y[i2] > 'Z' && y[i2] < 'a') || y[i2] >= 'z')
			{
				y[i2] = y[i2] - 26;
			}
		}
	}
	for(int i=0;i<17;i++)
		cout<<(char)x[i];
	putchar('\n');
	for(int i=0;i<38;i++)
		cout<<(char)y[i];
 
	return 0;
}