上学期看到2018年刑侦科推理试题，直觉告诉我可以嵌套N层for循环来解答，但后面还是用组合算法穷举出所有组合算到答案，嵌套太深不好搞啊。

从 0b00000000000000000000 自增到 0b11111111111111111111,每2bit表示一道题的答案，即: 0b00为A,0b01为B,0b10为C,0b11为D，

利用位移操作从左边高位依次读取每2bit的数，判断后不符合题意的组合就扔掉，最终肯定有一个组合答案符合。

下面我自己的解法

//

// Created by sys234 on 4/3/2018.

//

#include

#include

#include

#include

using namespace std;

const unsigned int A = 0;

const unsigned int B = 1;

const unsigned int C = 2;

const unsigned int D = 3;

vector getAnswerLine(unsigned int num) {

vector answer;

for (unsigned int j = 0; j < 20; j += 2) {

//offset 2bit

unsigned int x = ((num << (32 - 2 - j)) >> 30);

answer.push_back(x);

}

return answer;

}

bool isTheSame(int qnum, int qans) {

return qnum == qans;

}

bool notTheSame(int qnum, int qans) {

return qnum != qans;

}

bool isNear(int qans1, int qans2) {

return abs(qans1 - qans2) == 1;

}

unsigned int getAnsAppearTimes(const vector &alls, int max = -1, bool retAsTimes = false) {

map stat;

for (auto item : alls) {

int val = item;

++stat[val];

}

int minAns = 0;//supposing A

int qTimes = 0;

auto pair1 = stat.find(0);

minAns = 0;

qTimes = pair1->second;

for (int i = 1; i < 4; ++i) {

pair1 = stat.find(i);

if (qTimes > pair1->second && -1 == max) {

minAns = pair1->first;

qTimes = pair1->second;

continue;

}

if (qTimes < pair1->second && -1 != max) {

minAns = pair1->first;

qTimes = pair1->second;

continue;

}

}

if (retAsTimes == false) {

return minAns;

}

return qTimes;

}

unsigned int getAnsNumTimes(const vector &alls, int max = -1) {

return getAnsAppearTimes(alls, max, true);

}

char echoAns(int n) {

switch (n) {

case 0:

return ‘A‘;

case 1:

return ‘B‘;

case 2:

return ‘C‘;

case 3:

return ‘D‘;

default:

return ‘ ‘;

}

}

vector getAnswerLine(unsigned int num);

void combination() {

unsigned int ansMax = 0b11111111111111111111;//note: 20/2=10 ,2bit presents a answer;A=>00,B=>01,C=>10,D=>11

unsigned int ansMin = 0b0;

unsigned int i;

for (i = 0; i <= ansMax; ++i) {

vector one = getAnswerLine(i);

unsigned int q1 = one[0], q2 = one[1], q3 = one[2], q4 = one[3], q5 = one[4], q6 = one[5], q7 = one[6], q8 = one[7], q9 = one[8], q10 = one[9];

if (q1 == q2 && q1 == q3 && q1 == q4 && q1 == q5 && q1 == q6 && q1 == q7 && q1 == q8 && q1 == q9 && q1 == q10) {

//excluding all the same answer

continue;

}

if (isTheSame(q2, A) && notTheSame(q5, C)

|| isTheSame(q2, B) && notTheSame(q5, D)

|| isTheSame(q2, C) && notTheSame(q5, A)

|| isTheSame(q2, D) && notTheSame(q5, B)) {

continue;

//excluding 2nd

}

if (isTheSame(q3, A) && isTheSame(q3, q6)

|| isTheSame(q3, A) && isTheSame(q3, q2)

|| isTheSame(q3, A) && isTheSame(q3, q4)

//

|| isTheSame(q3, B) && isTheSame(q3, q6)

|| isTheSame(q3, B) && isTheSame(q6, q2)

|| isTheSame(q3, B) && isTheSame(q6, q4)

//

|| isTheSame(q3, C) && isTheSame(q3, q2)

|| isTheSame(q3, C) && isTheSame(q6, q2)

|| isTheSame(q3, C) && isTheSame(q2, q4)

//

|| isTheSame(q3, D) && isTheSame(q3, q4)

|| isTheSame(q3, D) && isTheSame(q6, q4)

|| isTheSame(q3, D) && isTheSame(q2, q4)

) {

continue;

//excluding 3rd

}

if (isTheSame(q4, A) && notTheSame(q1, q5)

|| isTheSame(q4, B) && notTheSame(q2, q7)

|| isTheSame(q4, C) && notTheSame(q1, q9)

|| isTheSame(q4, D) && notTheSame(q6, q10)) {

continue;

//excluding4

}

if (isTheSame(q5, A) && notTheSame(q8, A)

|| isTheSame(q5, B) && notTheSame(q4, B)

|| isTheSame(q5, C) && notTheSame(q9, C)

|| isTheSame(q5, D) && notTheSame(q7, D)) {

continue;

//excluding 5

}

if (isTheSame(q6, A) && (notTheSame(q2, q8) || notTheSame(q4, q8))

|| isTheSame(q6, B) && (notTheSame(q1, q8) || notTheSame(q6, q8))

|| isTheSame(q6, C) && (notTheSame(q3, q8) || notTheSame(q10, q8))

|| isTheSame(q6, D) && (notTheSame(q5, q8) || notTheSame(q9, q8))

) {

continue;

//excluding 6

}

int selectedLeastAlpha = getAnsAppearTimes(one, -1);

if (isTheSame(q7, A) && notTheSame(selectedLeastAlpha, C)

|| isTheSame(q7, B) && notTheSame(selectedLeastAlpha, B)

|| isTheSame(q7, C) && notTheSame(selectedLeastAlpha, A)

|| isTheSame(q7, D) && notTheSame(selectedLeastAlpha, D)

) {

continue;

//excluding 7

}

if (isTheSame(q8, A) && isNear(q1, q7)

|| isTheSame(q8, B) && isNear(q1, q5)

|| isTheSame(q8, C) && isNear(q1, q2)

|| isTheSame(q8, D) && isNear(q1, q10)

) {

continue;

//excluding 8

}

if (isTheSame(q9, A) && isTheSame(q1, q6) && isTheSame(q6, q5)

|| isTheSame(q9, A) && notTheSame(q1, q6) && notTheSame(q6, q5)

//

|| isTheSame(q9, B) && isTheSame(q1, q6) && isTheSame(q10, q5)

|| isTheSame(q9, B) && notTheSame(q1, q6) && notTheSame(q10, q5)

//

|| isTheSame(q9, C) && isTheSame(q1, q6) && isTheSame(q2, q5)

|| isTheSame(q9, C) && notTheSame(q1, q6) && notTheSame(q2, q5)

//

|| isTheSame(q9, D) && isTheSame(q1, q6) && isTheSame(q9, q5)

|| isTheSame(q9, D) && notTheSame(q1, q6) && notTheSame(q9, q5)

//

) {

continue;

//excluding 9th

}

int optionsCountMax = getAnsNumTimes(one, 1);

int optionsCountMin = getAnsNumTimes(one, -1);

int diff = optionsCountMax - optionsCountMin;

if (isTheSame(q10, A) && 3 != diff

|| isTheSame(q10, B) && 2 != diff

|| isTheSame(q10, C) && 4 != diff

|| isTheSame(q10, D) && 1 != diff

) {

continue;

//exclusing 10th

}

//print answer

for (int lastAns : one) {

cout << echoAns(lastAns)<

}

cout << endl;

}

}

int main() {

std::cout << clock() << std::endl;

combination();

std::cout << clock() << std::endl;

return 0;

}

解题中少了几个过滤条件，打印除算出好几行答案，差点以为我解法写错了，然后搜索到其他网友的答案，提有意思了，有嵌套多层for循环生成组合的。

有用4进制组合答案，多种语言好多技巧，学习了...

最终把缺少过滤条件判断加上去就OK了。