八皇后问题（递归ver)

因为老师提到所以去试试了

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#include<stdio.h>

#include<math.h>

#include<stdlib.h>//为了g++编译器里的abs函数

using namespace std;

int queenPlaces[92][8];//存放92种皇后棋子的摆放方法

int count = 0;

int board[8][8];//仿真棋盘

void putQueen(int ithQueen);//递归函数，每次摆好一个棋子

int main()

{

int n, i, j;

for (i = 0; i < 8; i++){//初始化

for (j = 0; j < 8; j++){

board[i][j] = -1;

}

for (j = 0; j < 92; j++) {

queenPlaces[j][i] = 0;

}

}

putQueen(0);//从第0个棋子开始摆放，运行的结果是将queenPlaces生成好

scanf("%d", &n);//不可以用scanf_s()

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

int ith;

scanf("%d", &ith);

for (j = 0; j < 8; j++) {

printf("%d", queenPlaces[ith - 1][j]);

}

printf("\n");

}

}

void putQueen(int ithQueen) {

int i, k, r;

if (ithQueen == 8)//如果摆满八个就计数加一

{

count++;

return;

}

for (i = 0; i < 8; i++)

{

if (board[i][ithQueen] == -1) {//摆放皇后

board[i][ithQueen] = ithQueen;

for (k = count; k < 92; k++)//没满则继续寻找

{

queenPlaces[k][ithQueen] = i + 1;

}

for (k = 0; k < 8; k++)

for (r = 0; r < 8; r++)

if (board[k][r] == -1 && (k == i || r == ithQueen || abs(k - i) == abs(r - ithQueen)))

board[k][r] = ithQueen;

putQueen(ithQueen + 1);//向下级递归

for (k = 0; k < 8; k++)//回溯，撤销控制范围

for (r = 0; r < 8; r++)

if (board[k][r] == ithQueen)

board[k][r] = -1;

}

}

}

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最后编译成功了，还挺开心的。qwq