pwnable.kr [collision]
2017-11-16 14:09
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题目
#include <stdio.h> #include <string.h> unsigned long hashcode = 0x21DD09EC; unsigned long check_password(const char* p){ int* ip = (int*)p; int i; int res=0; for(i=0; i<5; i++){ res += ip[i]; } return res; } int main(int argc, char* argv[]){ if(argc<2){ printf("usage : %s [passcode]\n", argv[0]); return 0; } if(strlen(argv[1]) != 20){ printf("passcode length should be 20 bytes\n"); return 0; } if(hashcode == check_password( argv[1] )){ system("/bin/cat flag"); return 0; } else printf("wrong passcode.\n"); return 0; }
WP:
题目的逻辑也很简单,要求输入20个字符,每四个字符为一组成为一个字节的数据,一共5组相加结果得到的值为hashcode就得到了flag.这个碰撞的概率很大,所以先尝试取一个字节的字符让他乘4,再用已知的hashcode减去得到最后4个字节数据,检查是否每个字节为可见字符。如果不满足重新尝试。我选的字符程序如下:#include<stdio.h> int main(){ int a=0x31283030; int hashcode=0x21dd09ec; int res = a+a+a+a; int t=0x5d3c492c; printf("0x%x\n",hashcode-res); printf("0x%x\n",t+res); return 0; }
最后得到的字符串为:”00(100(100(100(1,I<]”
输入这个即可得到flag
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