看了这么多位dalao的WP，写出了一份自己的爆破代码题目需要从：result[16]倒推二十次求出我们的Input[16]根据题设，有16个方程：cube[a][b][c] = d且这个64 * 64 * 64的立方体有一个很特殊的性质是：每一个64的方块都是 0 - 63 的不重复不遗漏的意味着，a,b,c,d四个数我们可以知三求一开始暴力：假设Input[16] = {x0,x1,……,x15}枚举x0，x1，根据cube数组以及result的结果，我们可以反向计算出来x2，即可以少爆破一个64的枚举这样算下来，需要爆破的就不是64 ^ 16，而是很少的几个值，剩下的值在数组中计算，再通过result检查即可代码如下：
#include <iostream>
#include <stdio.h>
using namespace std;
typedef unsigned char u8;
u8 cube[64][64][64];
u8 GetNumber1[64][64][64];
u8 GetNumber2[64][64][64];
u8 GetNumber3[64][64][64];
u8 result[16] = {0x14,0x22,0x1E,0x10,0x38,0x30,0x18,0x10,4,0x1A,0x24,8,2,0x26,0x38,0x2A};
void Print(int x){
printf("<%d>: ",x);
for(int i = 0; i < 16; i++)
printf("%02X ",result[i]);
char *sz = "abcdefghijklmnopqrstuvwxyz+-ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
for(int i = 0; i < 16; i++)
printf("%c",sz[result[i]]);
puts("");
}
void test(){
u8 x0,x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11,x12,x13,x14,x15;
for(x0 = 0; x0 < 64; x0++)
for(x1 = 0; x1 < 64; x1++){
x2 = GetNumber3[x0][x1][result[0]];
for(x3 = 0; x3 < 64; x3++){
x4 = GetNumber2[x3][result[1]][x0];
for(x5 = 0; x5 < 64; x5++){
x6 = GetNumber2[x5][result[2]][x0];
if (x6 == GetNumber2[x4][result[4]][x1]){
x7 = GetNumber2[x5][result[3]][x1];
x8 = GetNumber1[result[5]][x2][x3];
x9 = GetNumber1[result[6]][x2][x4];
x10 = GetNumber2[x7][result[7]][x3];
x12 = GetNumber2[x8][result[8]][x5];
x15 = GetNumber1[result[13]][x9][x10];
x11 = GetNumber2[x15][result[14]][x12];
x14 = GetNumber2[x11][result[11]][x9];
x13 = GetNumber2[x11][result[9]][x6];
if (GetNumber1[result[10]][x13][x7] == x12 &&
GetNumber1[result[12]][x8][x10] == x14 &&
GetNumber1[result[15]][x13][x14] == x15){
result[0] = x0;
result[1] = x1;
result[2] = x2;
result[3] = x3;
result[4] = x4;
result[5] = x5;
result[6] = x6;
result[7] = x7;
result[8] = x8;
result[9] = x9;
result[10] = x10;
result[11] = x11;
result[12] = x12;
result[13] = x13;
result[14] = x14;
result[15] = x15;
return;
}
}
}
}
}
}
int main(){
FILE *f = fopen("Escape.exe","rb");
fseek(f,0xe4f0,0);
fread(cube,64*64*64,1,f);
fclose(f);
for(int i = 0; i < 64; i++)
for(int j = 0; j < 64; j++)
for(int k = 0; k < 64; k++){
GetNumber1[cube[i][j][k]][j][k] = i;
GetNumber2[i][cube[i][j][k]][k] = j;
GetNumber3[i][j][cube[i][j][k]] = k;
}
Print(0);
for(int i = 1; i <= 0x500; i++){
test();
Print(i);
}
return 0;
}
