[AFCTF2018]MagicNum

题目

分析

double类型数据，转化成内部存储形式，然后再解码成字节码

#include <stdio.h>
#include <stdlib.h>

int main()
{
   double x;
   while(scanf("%llf",&x)){
     unsigned char *p;
   char buff[9];
   int i;

   p=(unsigned char *)&x;
   for (i=0; i<8; i++){
    itoa(*p++,buff,2);
    printf("%08s",buff);
   }
   printf("\n");
   }

   return 0;
}

#72065910510177138000000000000000.000000
c1=0b0110000101100110011000110111010000111101#这里把后面的0给去掉了，本来有64位
#71863209670811371000000.000000
c2=0b0110011001111011011100110110010100111101
#18489682625412760000000000000000.000000
c3=0b0110001101011111011010010111001100111101
#72723257588050687000000.000000
c4=0b0101111101100101011101100110010100111101
#4674659167469766200000000.000000
c5=0b0111001001111001011101110110100000111101
#19061698837499292000000000000000000000.000000
c6=0b0110010101110010011001010111110100111101
import binascii
flag=b''
for i in range(1,7):
    s='c'+str(i)
    print(binascii.unhexlify(hex(eval(s))[2:]))
    flag+=binascii.unhexlify(hex(eval(s))[2:])
print(flag)

把等号去掉就是flag了。