Mommy! I made a lotto program for my homework.
do you want to play?
ssh lotto@pwnable.kr -p2222 (pw:guest)
同样是一个小游戏,考查...细心程度。
源码如下:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <fcntl.h>
unsigned char submit[6];
void play(){
int i;
printf("Submit your 6 lotto bytes : ");
fflush(stdout);
int r;
r = read(0, submit, 6);
printf("Lotto Start!\n");
//sleep(1);
// generate lotto numbers
int fd = open("/dev/urandom", O_RDONLY);
if(fd==-1){
printf("error. tell admin\n");
exit(-1);
}
unsigned char lotto[6];
if(read(fd, lotto, 6) != 6){
printf("error2. tell admin\n");
exit(-1);
}
for(i=0; i<6; i++){
lotto[i] = (lotto[i] % 45) + 1; // 1 ~ 45
}
close(fd);
// calculate lotto score
int match = 0, j = 0;
for(i=0; i<6; i++){
for(j=0; j<6; j++){
if(lotto[i] == submit[j]){
match++;
}
}
}
// win!
if(match == 6){
system("/bin/cat flag");
}
else{
printf("bad luck...\n");
}
}
void help(){
printf("- nLotto Rule -\n");
printf("nlotto is consisted with 6 random natural numbers less than 46\n");
printf("your goal is to match lotto numbers as many as you can\n");
printf("if you win lottery for *1st place*, you will get reward\n");
printf("for more details, follow the link below\n");
printf("http://www.nlotto.co.kr/counsel.do?method=playerGuide#buying_guide01\n\n");
printf("mathematical chance to win this game is known to be 1/8145060.\n");
}
int main(int argc, char* argv[]){
// menu
unsigned int menu;
while(1){
printf("- Select Menu -\n");
printf("1. Play Lotto\n");
printf("2. Help\n");
printf("3. Exit\n");
scanf("%d", &menu);
switch(menu){
case 1:
play();
break;
case 2:
help();
break;
case 3:
printf("bye\n");
return 0;
default:
printf("invalid menu\n");
break;
}
}
return 0;
}
规则是输入一个 6 字节的字符串,与程序随机生成的 6 字节字符串比较( /dev/urandom 文件是 Linux 系统生成的char型随机数据,从这里读数据相当于产生不为空的随机字符流),相同则成功。并且由 lotto[i] = (lotto[i] % 45) + 1; // 1 ~ 45
可知 lotto 中字符的 ASCII 码为 1 到 45 。
ASCII 码表中只有 DEC 33 开始才是可见字符,所以需要输入的字符为 ASCII DEC 33 到 45 。
明确了规则还不够,因为从正常流程下猜中的几率几乎为 0 。
继续看代码,发现一个问题:
// calculate lotto score
int match = 0, j = 0;
for(i=0; i<6; i++){
for(j=0; j<6; j++){
if(lotto[i] == submit[j]){
match++;
}
}
}
这里的循环逻辑是,对于每 lotto[i] ,会和输入字串的每一个字符去比较,相等就使 match +1,而成功需要match = 6 。所以只要输入的字串为相同的6个字符,6个字符里有一个中了,那就中了 :P
又由于是随机产生的 lotto 串,这里只要重复尝试一个字串就行了:
……
Submit your 6 lotto bytes : ''''''
Lotto Start!
bad luck...
- Select Menu -
1. Play Lotto
2. Help
3. Exit
1
Submit your 6 lotto bytes : ''''''
Lotto Start!
sorry mom... I FORGOT to check duplicate numbers... :(
至于需要尝试几次就完全看运气了。
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