最近看到很火的那个2018年刑侦推理题，很多程序员都用来编程暴力破解 感觉挺好玩，就自己也来用c++试一试 （正在学校机房上实验课的我，只能对着发白的codeblocks撸码）
先上结果
BCACACDABA /0.086秒
代码
其实大家的方法都一样啦 就是穷举，判断。 因为写之前看到知乎上大佬把10个题目都放在一个判断语句里面判断了 觉得正合我意，便也只用一个if判断句
#include <iostream>
#include <math.h>
using namespace std;
bool test(int q[],int an[],int temp);
int main() {
int issue[10]={0};
for (int i = 0; i < 1048576; i++) {
for(int j=0;j<10;j++){
int k=i/pow(4,j);
issue[j]=k % 4;
}
int an[4] = { 0 }, min = 10, max = 0;
for (int j = 0; j < 10; j++) an[issue[j]]++;
for (int j = 0; j < 4; j++){
if(max<an[j])max=an[j];
if(min>an[j])min=an[j];
}
if(test(issue,an,max-min)) {
for (int j = 0; j < 10; j++)
cout<<"第"<<j<<"题："<<char(A + issue[j])<<endl;
break;
}
}
}
bool test(int q[10],int an[4],int temp)
{
if ((q[1] == 0 && q[4] == 2 ||q[1] == 1 && q[4] == 3 ||
q[1] == 2 && q[4] == 0 ||q[1] == 3 && q[4] == 1) &&
(q[2] == 0 && q[2] != q[5] && q[2] != q[1] && q[2] != q[3] ||
q[2] == 1 && q[5] != q[2] && q[5] != q[1] && q[5] != q[3] ||
q[2] == 2 && q[1] != q[2] && q[1] != q[5] && q[1] != q[3] ||
q[2] == 3 && q[3] != q[2] && q[3] != q[5] && q[3] != q[1]) &&
(q[3] == 0 && q[0] == q[4] ||q[3] == 1 && q[1] == q[6] ||
q[3] == 2 && q[0] == q[8] ||q[3] == 3 && q[5] == q[9]) &&
(q[4] == 0 && q[7] == 0 ||q[4] == 1 && q[3] == 1 ||
q[4] == 2 && q[8] == 2 ||q[4] == 3 && q[6] == 3) &&
(q[5] == 0 && q[1] == q[7] && q[3] == q[7] ||
q[5] == 1 && q[0] == q[7] && q[5] == q[7] ||
q[5] == 2 && q[2] == q[7] && q[9] == q[7] ||
q[5] == 3 && q[4] == q[7] && q[8] == q[7]) &&
(q[6] == 0 && an[2] < an[0] && an[2] < an[1] && an[2] < an[3] ||
q[6] == 1 && an[1] < an[0] && an[1] < an[2] && an[1] < an[3] ||
q[6] == 2 && an[0] < an[1] && an[0] < an[2] && an[0] < an[3] ||
q[6] == 3 && an[3] < an[0] && an[3] < an[1] && an[3] < an[2]) &&
(q[7] == 0 && (q[0] - q[6])*(q[0] - q[6]) != 1 ||
q[7] == 1 && (q[0] - q[4])*(q[0] - q[4]) != 1 ||
q[7] == 2 && (q[0] - q[1]) *(q[0] - q[1])!= 1 ||
q[7] == 3 && (q[0] - q[9]) *(q[0] - q[9])!= 1) &&
(q[8] == 0 && (q[0] == q[5]) != (q[5] == q[4]) ||
q[8] == 1 && (q[0] == q[5]) != (q[9] == q[4]) ||
q[8] == 2 && (q[0] == q[5]) != (q[1] == q[4]) ||
q[8] == 3 && (q[0] == q[5]) != (q[8] == q[4])) &&
(q[9] == 0 && temp == 3 ||q[9] == 1 && temp == 2 ||
q[9] == 2 && temp == 4 ||q[9] == 3 && temp == 1))
return true;
else return false;
}
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