GFSJ0489-【open-source】
约束求解
源码
一共三个约束
#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[]) {
if (argc != 4) {
printf("what?\n");
exit(1);
}
unsigned int first = atoi(argv[1]);
if (first != 0xcafe) {
printf("you are wrong, sorry.\n");
exit(2);
}
unsigned int second = atoi(argv[2]);
if (second % 5 == 3 || second % 17 != 8) {
printf("ha, you won't get it!\n");
exit(3);
}
if (strcmp("h4cky0u", argv[3])) {
printf("so close, dude!\n");
exit(4);
}
printf("Brr wrrr grr\n");
unsigned int hash = first * 31337 + (second % 17) * 11 + strlen(argv[3]) - 1615810207;
printf("Get your key: ");
printf("%x\n", hash);
return 0;
}
exp
first = 0xcafe
second = 8
third = "h4cky0u"
key = first * 31337 + (second % 17) * 11 + len(third) - 1615810207
print(hex(key)[2:])
也可以这样解
gcc编译一下 然后传参
flag
c0ffee
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