Perlis教授折磨脑子的家庭作业

#include
#include
#define NUM 5
void repeat(int num);
void print();
char s[NUM + 1];
int k = 0;

struct list {
	int token;
	char load;
};
struct list  qsort[NUM + 1] = {0};
char copy[NUM + 1];
int main() {
	
	scanf("%s", s);
	int num;
	num = strlen(s);
	
	repeat(num);

	return 0;
}

void repeat(int num) {
	
	for (int i = num, j = NUM; i > 0; i--)
	{
		while((j-1)>=0)
			while (qsort[--j].token == 0 && j>=0)
			{
				qsort[NUM - num].load = s[j];
				qsort[j].token = 1;

				if(num-1>0)
				repeat(num - 1);
			}

	}
	
	qsort[NUM].load = '';
			
	for (int i = NUM ; i >= 0; i--)//如不想留，可根据自身情况进行删除改进
		copy[i] = qsort[i].load;
				
	for (int i = num, j = NUM - 1; i >= 0; i--, j--)
		for (int m = NUM - 1; m >= 0; m--)
			if (s[m] == copy[j])
				qsort[m].token = 0;
	
	if (num == 1)
		print();
	return;
}
void print() {
	static int k=0;
	printf("%s ", copy);
	k++;
	if (k % 10 == 0)
		printf("n");
}

解释：在8×8格的国际象棋上摆放8个皇后，使其不能互相攻击，即任意两个皇后都不能处于同一行、同一列或同一斜线上，问有多少种摆法。

题解：由于八个棋子不同行不同列，所以总会有棋子在第1,2,3,4,5,6,7,8行，可将问题类化为将1 2 3 4 5 6 7 8排序，求有多少中排法？
所以将第一题中的NUM宏定义为8，字符串输入为“12345678”即可求解

#include
#include
#define NUM 8
void repeat(int num);
void print();
char s[NUM + 1];
int k = 0;

struct list {
	int token;
	char load;
};
struct list  qsort[NUM + 1] = {0};
char copy[NUM + 1];
int main() {
	
	scanf("%s", s);
	int num;
	num = strlen(s);
	
	repeat(num);

	return 0;
}

void repeat(int num) {
	
	for (int i = num, j = NUM; i > 0; i--)
	{
		while((j-1)>=0)
			while (qsort[--j].token == 0 && j>=0)
			{
				qsort[NUM - num].load = s[j];
				qsort[j].token = 1;

				if(num-1>0)
				repeat(num - 1);
			}

	}
	
	qsort[NUM].load = '';
			
	for (int i = NUM ; i >= 0; i--)//如不想留，可根据自身情况进行删除改进
		copy[i] = qsort[i].load;
				
	for (int i = num, j = NUM - 1; i >= 0; i--, j--)
		for (int m = NUM - 1; m >= 0; m--)
			if (s[m] == copy[j])
				qsort[m].token = 0;
	
	if (num == 1)
		print();
	return;
}
void print() {
	static int k=0;
	printf("%s ", copy);
	k++;
	if (k % 10 == 0)
		printf("n");
}

#include
int getnum();
void list(int n);
void print(int i);
int main() {
	int num;
	
	num = getnum();
	
	list(num);
	return 0;
}
int getnum() {
	int num;
	scanf("%d", &num);
	return num;
}

void list(int n) {
	for (int i = 2; i <= n; i++) {
		int token = 1;
		for (int j = 2; j <= i / 2; j++) {
			
			if (i == 2)
			{
				break;
			}
			
			if (i % j == 0) {
				token = 0;
				break;
			}
		}
		if(token ==1)
			print(i);
	}
}

void print(int i) {
	printf("%d ",i);
}

#include

#define LINE1 3
#define ROW1 2
#define LINE2 2
#define ROW2 3

void input1();
void input2();
void multiple();

int matrix1[LINE1][ROW1];
int matrix2[LINE2][ROW2];
int main() {
	
    
    input1();
    
    input2();
    
    if (ROW1 == LINE2) 
        multiple();
    else printf("第一个矩阵的列数不等于第二个矩阵的行数，矩阵不能进行乘法n");

}

void input1() {
    printf("输入%d行%d列的matrix1矩阵:n", LINE1, ROW1);
    
    for (int i = 0; i < LINE1; i++) {
        for (int j = 0; j < ROW1; j++) {
            scanf("%d", &matrix1[i][j]);
        }
    }
}

void input2() {
    printf("输入%d行%d列的matrix2矩阵:n", LINE2, ROW2);
    for (int i = 0; i < LINE2; i++) {
        for (int j = 0; j < ROW2; j++) {
            scanf("%d", &matrix2[i][j]);
        }
    }
}

void multiple() {
    printf("matrix1矩阵与matrix2矩阵乘积为:n");
    for (int i = 0; i < LINE1; i++) {//最终获得行数
        for (int j = 0, sum = 0; j < ROW2; j++) {//最终获得列数

            for (int k = 0; k < LINE2; k++)
                sum += matrix1[i][k] * matrix2[k][j];
            
            printf("%3d ", sum);
            sum = 0;
        }
        printf("n");
    }
}