# 家教解答
題目：
https://docs.google.com/document/d/1xcLkz2zezrbLb0pE_CQ8v7b5vDbvAWhlKsktEUtHDpw/edit?usp=sharing

## Week 1 question

### Q1

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main(){
    char ID[10];
    float Hcm;
    float W;
    float Hm;
    float BMI;

    printf("Enter ID: ");
    scanf("%s", ID);
    printf("Enter weight: ");
    scanf("%f", &W);
    printf("Enter height(cm): ");
    scanf("%f", &Hcm);

    Hm = Hcm / 100;
    BMI = W / pow(Hm,2);
    printf("%s Your BMI: %.1f", ID, BMI);
}

### Q2

#include <stdio.h>
#include <stdlib.h>

int main(){
    int n;
    int ans = 1;

    printf("Enter n: ");
    scanf("%d", &n);

    if (n == 0){
    	printf("0! is 1");
    }
    else if (n < 0){
    	printf("number can't be less than 0");
    }else{
    	int tmp = n;
    	while (tmp > 0){
            ans *= tmp;
            tmp--;
    	}
    	printf("%d! is %d", n, ans);
    }
}

### Q3: a006. 一元二次方程式

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main(){
    int a, b, c;
    int ans1, ans2;

    printf("請輸入a b c，中間使用空白區隔");
    scanf("%d %d %d", &a, &b, &c);

    int num_in_sqrt = pow(b,2) - 4*a*c;

    if (num_in_sqrt < 0){
    	printf("No real root");
    }else{
    	ans1 = (sqrt(num_in_sqrt) - b) / 2 * a;
    	ans2 = (-sqrt(num_in_sqrt) - b) / 2 * a;

    	if (ans1 == ans2){
            printf("Two same roots x=%d", ans1);
        }else{
            printf("Two different roots x1=%d , x2=%d", ans1, ans2);
        }
    }
}

### Q4: 10進位轉16進位

#include <stdio.h>
#include <stdlib.h>

char buffer[33]; //用於存放轉換好的十六進位制字串，可根據需要定義長度

char* IntToHex(int aa)
{
    sprintf(buffer, "%x", aa);
    return buffer;
}

int main ()
{
    int num;
    char* hex_str;
    printf ("Enter a number: ");
    scanf ("%d",&num);
    printf("%x", num);
    hex_str = IntToHex (num);
    printf ("Hexadecimal number: %sH\n", hex_str);
    return 0;
}

## Q5: Remove Duplicates from Sorted Array
### 題目敘述
input 會給你一個排序過後的陣列，實作一個 function 把重複的值剔除(in-place 移除，不分配額外的空間)，使每個元素都只出現一次，function 回傳陣列的新長度。

### 解法思路
用兩個index（i比較慢 & j比較快），如遇到重複的則i不會動，直到index j跳脫duplicate value的區域，如此確保index i所處理的資料均無重複

int removeDuplicates(int* nums, int numsSize) {
    if (numsSize == 0) return 0;

    int i = 0;
    for (int j = 1; j < numsSize; j++) {
        if (nums[j] != nums[i]) {
            i++;
            nums[i] = nums[j];
        }
    }
    return i + 1;
}

## Week 2 question

### Q1
略

### Q2
Leetcode - Search Insert Position
法1: 暴力法

#include <stdio.h>
#include <stdlib.h>

int searchInsert(int* nums, int numsSize, int target) {
    int i;
    for(i=0;i<numsSize;i++)
    {
        if(nums[i]>=target)
            return i;
    }
    return i;
}

int main ()
{
    int nums[] = {1,3,5,6};
    int numsSize = sizeof(nums) / sizeof(int);
    int target;
    printf("Enter a num: ");
    scanf("%d", &target);

    int result = searchInsert(nums, numsSize, target);
    printf("Inserted index is %d", result);
}

法2: 二分法

#include <stdio.h>
#include <stdlib.h>

int searchInsert(int* nums, int numsSize, int target) {
    int left = 0, right = numsSize - 1, mid = 0;
    while(left <= right) {
        mid = (left + right) / 2;
        if(nums[mid] == target)
            return mid;
        else if(nums[mid] > target) {
            right = mid - 1;
        }
        else
            left = mid + 1;
    }
    return left;
}

int main ()
{
    int nums[] = {1,3,5,6};
    int numsSize = sizeof(nums) / sizeof(int);
    int target;
    printf("Enter a num: ");
    scanf("%d", &target);

    int result = searchInsert(nums, numsSize, target);
    printf("Inserted index is %d", result);
}

### Q3
Leetcode - Roman to Integer
#### 思路：
使用一個map來儲存羅馬符號跟數字之間的對應關係，在一般的情況下(ex. III, VI)，可以直接將羅馬符號轉換成數字。
不過如果出現IV，XC這種組合，就要另外處理，這種組合的特色是後面的符號會大於前面的符號，因此一次讀兩個羅馬符號來找出這種組合。 一次讀兩個羅馬數字，如果第二個數字(b)比第一個(a)大，b-a，b <= a，b+a。

#include<stdio.h>
#include<string.h>

int toNumber(char ch) {  
    switch (ch) {  
        case 'I': return 1;  
        case 'V': return 5;  
        case 'X': return 10;  
        case 'L': return 50;  
        case 'C': return 100;  
        case 'D': return 500;  
        case 'M': return 1000;
        default: return -1;   
    }   
} 

int romanToInt(char* s) {
    int total = 0;
    int i = 0;
    int a = 0;
    int b = 0;
    
    while(s[i] != '\0')
    {
        a=toNumber(s[i]);
        b=toNumber(s[i+1]);
        if(a >= b) 
        {
            total += a;
            i++;
        }
        else
        {
            total = total + b - a;
            i += 2;
        }
    }
    return total;
}
 
int main(){
    char roman_Number[1000];
    printf("Enter any roman number (Valid digits are I, V, X, L, C, D, M):  \n");
    scanf("%s",roman_Number);
    int sum = romanToInt(roman_Number);
    if (sum != -1)
    	printf("Its decimal value is : %d",sum);
    else
    	printf("Wrong input");
    return 0;
}

### Q4
Leetcode - Excel Sheet Column Number
#### 思路
因為A的ASC II碼是65，那麼A-64=1。因此ABCD...等價於1，2，3，4...，Z等價於26，滿26進1。

接下來就是分別計算各個字母再相加就行了

#include<stdio.h>
#include<string.h>

int titleToNumber(char* s){
    int len = strlen(s);
    int output = 0;
    for(int i = 0; i < len; i++)
    {
        int temp = *(s+i) - 64; //臨時變量存放當前位上的值  *(s+i) = s[i]
        for(int j = 0; j < len-i-1; j++)
        {
            temp *= 26; //當前位的值乘上此位的權重，即乘以26的乘方
        }
        output += temp; //把計算的權重值逐位相加
    }
    return output;
}
 
int main(){
    char title[1000];
    printf("Enter column title:\n");
    scanf("%s",title);
    int sum = titleToNumber(title);
    printf("Its decimal value is : %d",sum);
    return 0;
}

## HW1
題目：
https://acm.cs.nthu.edu.tw/problem/12187/
解答：

#include <stdio.h>
#include <stdlib.h> // for atoi
#include <string.h>
#include <stdbool.h>
#include <limits.h> 

int main(void)
{
    int total_instruction_num;
    scanf("%d", &total_instruction_num);
    getchar();  // 吃掉換行字符

    bool isFirst = true;
    int ans = 0;
    int total_num = 0;

    for (int i = 0; i < total_instruction_num; i++){
        char s[50];
        fgets(s, sizeof(s), stdin); // scanf碰到字串有空格會直接斷掉
        // gets(s);  // 不安全的function，以fgets代替
        char *sub_str = strtok(s, " ");  // 依照空白做切割
        if (strcmp(sub_str, "sum") == 0){
            sub_str = strtok(NULL, " ");  // 把它想成取下一組就好
            while (sub_str != NULL) {
                if (isFirst){
                    isFirst = false;
                    total_num += atoi(sub_str);  // atoi = alphbet to integer
                }
                else{
                    int num = atoi(sub_str);
                    ans += num;
                }
                sub_str = strtok(NULL, " ");
            }
        }
        else if (strcmp(sub_str, "min") == 0){
            sub_str = strtok(NULL, " ");
            int min = INT_MAX;
            while (sub_str != NULL) {
                int num = atoi(sub_str);
                if (num < min){
                    min = num;
                }
                sub_str = strtok(NULL, " ");
            }
            ans += min;
            total_num++;
        }
        else{
            sub_str = strtok(NULL, " ");
            int max = INT_MIN;
            while (sub_str != NULL) {
                int num = atoi(sub_str);
                if (num > max){
                    max = num;
                }
                sub_str = strtok(NULL, " ");
            }
            ans += max;
            total_num++;
        }
        isFirst = true;
    }

    printf("%d %d", ans, total_num);
     
    return 0;    
}

## HW2
題目：
讀檔，兩個字串相乘
解法：

#include <stdio.h> 

// int charToint(char str){
// 	return (int)str - '0';
// }

int main() 
{ 
	// region: 讀檔&生成矩陣
	char InputFileName[] = "input.txt"; 
	char OutputFileName[] = "../output.txt"; 
	FILE *fp; 
	FILE *out_fp;
	char StrLine[1024];             //每行最大讀取的字元數
	int m,n,k;

	if((fp = fopen(InputFileName,"r")) == NULL) //判斷檔案是否存在及可讀
	{ 
		printf("error!"); 
		return -1; 
	} 

	fscanf(fp, "%d %d %d", &m, &n, &k);
	int a[m][n],b[n][k],ans[m][k];

	while (!feof(fp)) 
	{ 
		printf("---preprocess---\n");
		for(int i=0;i<m;i++)
	    {
	    	for(int j=0;j<n;j++)
	    	{  
	    		fscanf(fp, "%d", &a[i][j]);
	    		printf("%d ", a[i][j]);
	    	}
	    	printf("a\n");
	    }

	    for(int i=0;i<n;i++)
	    {
	    	for(int j=0;j<k;j++)
	    	{  
	    		fscanf(fp, "%d", &b[i][j]);
	    		printf("%d ", b[i][j]);
	    	}
	    	printf("b\n");
	    }
	    printf("---preprocess done---\n");

	    if((out_fp = fopen(OutputFileName,"w")) == NULL) //判斷檔案是否存在及可讀
		{ 
			printf("error!"); 
			return -1; 
		}

	    for(int i=0;i<m;i++)
	    {
	    	for(int j=0;j<k;j++){
	    		ans[i][j] = 0;
	    		for(int q=0;q<n;q++){
		        	ans[i][j] += a[i][q] * b[q][j]; 
		        }
		        fprintf(out_fp, "%d ", ans[i][j]);
	    	}
	    	fprintf(out_fp, "\n"); 
	    }

	



		/// region: 因為有負數，所以不能這樣preprocess
		// for(int i=0;i<m;i++)
	 //    {
	 //    	fgets(StrLine,1024,fp);  //讀取一行
	 //    	for(int j=0;j<n;j++)
	 //    	{  
	 //    		a[i][j] = charToint(StrLine[2*j]);  
	 //    		printf("%d ", a[i][j]);
	 //    	}
	 //    	printf("\n");
	 //    }

	 //    for(int i=0;i<n;i++)
	 //    {
	 //    	fgets(StrLine,1024,fp);  //讀取一行
	 //    	for(int j=0;j<k;j++)
	 //    	{  
	 //    		b[i][j] = charToint(StrLine[2*j]);    
	 //    		printf("%d ", b[i][j]);
	 //    	}
	 //    	printf("\n");
	 //    }
		///
	} 
	

	fclose(fp);                     //關閉檔案
	// endregion
	return 0; 
}

## 題庫
a022: 迴文

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* get_reverse(char* src){
    int n = strlen(src);
    char* src_reverse = malloc(n);
    for (int i = 0; i < n; i++){
        src_reverse[n-i-1] = src[i];
    }

    return src_reverse;
}

int main(){
    char src[1000];
    while (scanf("%s", src) != EOF){
        char* src_reversed = get_reverse(src);
        if (strcmp(src, src_reversed) == 0){
            printf("yes\n");
        }else{
            printf("no\n");
        }
    }
}

a015: 矩陣的翻轉

#include <stdio.h>
#include <stdlib.h>

int main(){
    int row;
    int col;

    while (scanf("%d %d", &row, &col) != EOF){
        int input_matrix[row][col];
        int output_matrix[col][row];

        for (int i = 0; i < row; i++){
            for (int j = 0; j < col; j++){
                scanf("%d", &input_matrix[i][j]);
                output_matrix[j][i] = input_matrix[i][j];
            }
        }

        for (int i = 0; i < col; i++){
            for (int j = 0; j < row; j++){
                printf("%d ", output_matrix[i][j]);
            }
            printf("\n");
        }
    }
}

## 補充
讀取任意輸入數量的數字
https://www.itread01.com/content/1546317753.

# 期中考
Q1: oj 12216

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int m,n,k;
    int s=1;

    scanf("%d %d %d",&m,&n,&k);  //從INPUT 讀第一行
    int a[m][n];
    int b[n][k];
    int ans[m][k];

    // *** here ***
    for(int i=0;i<m;i++)
    {
        for(int j=0;j<n;j++)
        {  
            scanf("%d", &a[i][j]);
        }
    }

    for(int i=0;i<n;i++)
    {
        for(int j=0;j<k;j++)
        {  
            scanf("%d", &b[i][j]);
        }
    }
    // *** here ***

    for(int i=0;i<m;i++) {         //計算a*b矩陣=ans
        for(int j=0;j<k;j++) {
            ans[i][j]=0;
            for(int q=0;q<n;q++) {
                ans[i][j] += a[i][q]*b[q][j];
            }
        }
    }

    for(int i=0;i<m;i++) {     //寫出a*b矩陣
        for(int j=0;j<k;j++) {
            if(s==1) {
                printf("%d",ans[i][j]);
                s=0;
            }
            else if(s==0) {  // *** here ***
                printf(" %d",ans[i][j]);
            }
        }
        s=1;
        if(i<m-1) {
            printf("\n");
        }
    }
}

# 作業
postfix calculation

#include <stdio.h>

int stack[100];
int top=-1;

void push(int x){
    top = top + 1;
    stack[top]=x;
}

int pop(){
    int res;
    res=stack[top--];
    return res;
}

int main(){
    char eq[100];
    int i=0,num,n1,n2,n3;
    scanf("%s",eq);
    while(eq[i]!=NULL)    {
        if(isdigit(eq[i]))        
        {
            num=eq[i]-48;
            push(num);
        }        
        else        
        {
            n1=pop();
            n2=pop();
            switch(eq[i])            
            {
                case'+':
                n3=n1+n2; push(n3);
                break;
                case'-':
                n3=n2-n1;  push(n3);
                break;
                case'*':
                n3=n1*n2;  push(n3);
                break;
                case'/':
                n3=n2/n1;  push(n3);
                break;
            }
        }
        i++;
    }
    printf("%d",n3);
    return 0;
}

infix calculation

#include <stdio.h>
#include <ctype.h>

int stack[100];
int top=-1;

void push(int x){
    top = top + 1;
    stack[top]=x;
}

int pop(){
    int res;
    res=stack[top--];
    return res;
}

int priority(char op) { 
    switch(op) { 
        case '+': case '-': return 1;
        case '*': case '/': return 2;
        default:            return 0;
    } 
} 

void inToPostfix(char* infix, char* postfix) { 
    char local_stack[80] = {'\0'};
    int i, j, local_top;
    for(i = 0, j = 0, local_top = 0; infix[i] != '\0'; i++){
        switch(infix[i]) { 
            case '(':              // 運算子堆疊 
                local_stack[++local_top] = infix[i]; 
                break; 
            case '+': case '-': case '*': case '/': 
                while(priority(local_stack[local_top]) >= priority(infix[i])) { 
                    postfix[j++] = local_stack[local_top--];
                } 
                local_stack[++local_top] = infix[i]; // 存入堆疊 
                break; 
            case ')': 
                while(local_stack[local_top] != '(') { // 遇 ) 輸出至 ( 
                    postfix[j++] = local_stack[local_top--];
                } 
                local_top--;  // 不輸出 ( 
                break; 
            default:  // 運算元直接輸出 
                postfix[j++] = infix[i];
        }
    } 
    while(local_top > 0) { 
        postfix[j++] = local_stack[local_top--];
    }
} 

int main(){
    char infix[80] = {'\0'}; 
    char postfix[80]= {'\0'}; 

    // printf("中序運算式："); 
    scanf("%s", infix); 
    inToPostfix(infix, postfix);
    // printf("%s", postfix);

    // char postfix[100];
    int i=0,num,n1,n2,n3;
    while(postfix[i]!='\0')    {
        if(isdigit(postfix[i]))        
        {
            num=postfix[i]-48;
            push(num);
        }        
        else        
        {
            n1=pop();
            n2=pop();
            switch(postfix[i])            
            {
                case'+':
                n3=n1+n2; push(n3);
                break;
                case'-':
                n3=n2-n1;  push(n3);
                break;
                case'*':
                n3=n1*n2;  push(n3);
                break;
                case'/':
                n3=n2/n1;  push(n3);
                break;
            }
        }
        i++;
    }
    printf("%d",n3);
    return 0;
}

Week15

#include <stdio.h>
#include <stdbool.h>  
#include <stdlib.h>
#include <string.h>

#define M 6
#define N 6

typedef struct node_t {
    int x, y; 
    int dir; 
}node;

#define MAXSIZE 100 /*定義最大堆疊容量*/
node s[MAXSIZE];  //堆疊的陣列宣告 
int top=-1;		//堆疊的頂端

int isempty() {
   if(top == -1)
      return 1;
   else
      return 0;
}
   
int isfull() {
   if(top == MAXSIZE)
      return 1;
   else
      return 0;
}

node peek() {
   return s[top];
}

node pop() {
   node data;
	
   if(!isempty()) {
      data = s[top];
      top = top - 1;   
      return data;
   } else {
      printf("Could not retrieve data, Stack is empty.\n");
   }
}

void push(node data) {
   if(!isfull()) {
      top = top + 1;   
      s[top] = data;
   } else {
      printf("Could not insert data, Stack is full.\n");
   }
}
  
// Coordinates of food 
int fx, fy; 
bool unvisited[M][N]; 
  
bool isReachable(int maze[M][N]) 
{ 
    // Initially starting at (0, 0). 
    int i = 0, j = 0; 
      
    node temp = { i, j, 0 };  
      
    push(temp); 
  
    while (!isempty()) { 
  
        // Pop the top node and move to the 
        // left, right, top, down or retract  
        // back according the value of node's 
        // dir variable. 
        temp = peek(); 
        int d = temp.dir; 
        i = temp.x, j = temp.y; 
  
        // Increment the direction and 
        // push the node in the stack again. 
        temp.dir++; 
        pop(); 
        push(temp); 
  
        // If we reach the Food coordinates 
        // return true 
        if (i == fx && j == fy) { 
            return true; 
        } 
  
        // Checking the Up direction. 
        if (d == 0) { 
            if (i - 1 >= 0 && maze[i - 1][j] &&  
                                    unvisited[i - 1][j]) { 
                node temp1 = { i - 1, j, 0 }; 
                unvisited[i - 1][j] = false; 
                push(temp1); 
            } 
        } 
  
        // Checking the left direction 
        else if (d == 1) { 
            if (j - 1 >= 0 && maze[i][j - 1] &&  
                                    unvisited[i][j - 1]) { 
                node temp1 = {i, j - 1, 0 };
                unvisited[i][j - 1] = false; 
                push(temp1); 
            } 
        } 
  
        // Checking the down direction 
        else if (d == 2) { 
            if (i + 1 < M && maze[i + 1][j] &&  
                                    unvisited[i + 1][j]) { 
                node temp1 = {i + 1, j, 0 };
                unvisited[i + 1][j] = false; 
                push(temp1); 
            } 
        } 
        // Checking the right direction 
        else if (d == 3) { 
            if (j + 1 < N && maze[i][j + 1] &&  
                                    unvisited[i][j + 1]) { 
                node temp1 = {i, j + 1, 0 };
                unvisited[i][j + 1] = false; 
                push(temp1); 
            } 
        } 
  
        // If none of the direction can take  
        // the rat to the Food, retract back  
        // to the path where the rat came from. 
        else { 
            unvisited[temp.x][temp.y] = true; 
            pop(); 
        } 
    } 
  
    // If the stack is empty and 
    // no path is found return false. 
    return false; 
} 

// Driver code 
int main() 
{ 
    // Initially setting the unvisited 
    // true = unvisited
    memset(unvisited, true, sizeof(unvisited)); 

    char InputFileName[] = "input.txt"; 
    FILE *fp; 
    char StrLine[1024];    //每行最大讀取的字元數

    if((fp = fopen(InputFileName,"r")) == NULL) //判斷檔案是否存在及可讀
	{ 
		printf("error!"); 
		return -1; 
	} 

    // Maze matrix 
    int maze[M][N] = {0}; 

    for(int i=0;i<M;i++)
    {
		for(int j=0;j<N;j++)
		{  
			fscanf(fp, "%d", &maze[i][j]);
		}
	}
  
  	fscanf(fp, "%d %d", &fx, &fy);
  
    if (isReachable(maze))
        printf("Yes\n"); 
    else
    	printf("No\n"); 
          
    for (int i = 0; i <= top; i++){
    	printf("(%d %d) ", s[i].x, s[i].y);
    }

    return 0; 
} 

## final
更改成可重複遊玩版本

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 10

int chessboard[N + 1][N + 1] = { 0 };

int whoseTurn = 0;

void initGame();
void printChessboard();
int playChess();
int judge(int, int);

void play(){
    initGame();
    while (1)
    {
        ++whoseTurn;
        int isEnd = playChess();
        if (isEnd != 0) break;
    }
}

int main()
{
    while (1){
        play();
        getchar();
        printf("Continue? y or n");
        char again;  
        scanf("%c", &again);
        if (again != 'y'){
            printf("%c", again);
            break;
        }
    }
    return 0;
}

void initGame(void)
{
    whoseTurn = 0;
    memset(chessboard, 0, sizeof(chessboard));

    system("cls");
    printChessboard();
}

void printChessboard()
{
    int i, j;

    for (i = 0; i <= N; i++)
    {
        for (j = 0; j <= N; j++)
        {
            if (0 == i)
                printf("%3d", j);
            else if (j == 0)
                printf("%3d", i);
            else if (1 == chessboard[i][j])
                printf("  O");
            else if (2 == chessboard[i][j])
                printf("  X");
            else
                printf("  *");
        }
        printf("\n");
    }
}

int playChess(void)
{
    int i, j;

    if (1 == (whoseTurn & 1))
    {
        printf("Turn to player 1, please input the position:");
        scanf("%d %d", &i, &j);

        while (chessboard[i][j] != 0)
        {
            printf("This position  has been occupied, please input the position again:");
            scanf("%d %d", &i, &j);
        }

        chessboard[i][j] = 1;
    }
    else
    {
        printf("Turn to player 2, please input the position:");
        scanf("%d %d", &i, &j);

        while (chessboard[i][j] != 0)
        {
            printf("This position  has been occupied, please input the position again:");
            scanf("%d %d", &i, &j);
        }

        chessboard[i][j] = 2;
    }

    system("cls");
    printChessboard();

    if (judge(i, j))
    {
        if (1 == (whoseTurn & 1))
        {
            printf("Winner is player 1!\n");
            return 1;
        }
        else
        {
            printf("Winner is player 2!\n");
            return 2;
        }
    }
    return 0;
}

int judge(int x, int y)
{
    int i, j;
    int t = 2 - (whoseTurn & 1);

    for (i = x - 4, j = y; i <= x; i++)
    {
        if (i >= 1 && i <= N && t == chessboard[i][j] && t == chessboard[i + 1][j] && t == chessboard[i + 2][j] && t == chessboard[i + 3][j] && t == chessboard[i + 4][j])
            return 1;
    }
    for (i = x, j = y - 4; j <= y; j++)
    {
        if (j >= 1 && j <= N && t == chessboard[i][j] && t == chessboard[i][j + 1] && t == chessboard[i][j + 2] && t == chessboard[i][j + 3] && t == chessboard[i][j + 4])
            return 1;
    }
    for (i = x - 4, j = y - 4; i <= x, j <= y; i++, j++)
    {
        if (i >= 1 && i <= N && j >= 1 && j <= N && t == chessboard[i][j] && t == chessboard[i + 1][j + 1] && t == chessboard[i + 2][j + 2] && t == chessboard[i + 3][j + 3] && t == chessboard[i + 4][j + 4])
            return 1;
    }
    for (i = x + 4, j = y - 4; i >= 1, j <= y; i--, j++)
    {
        if (i >= 1 && i <= N && j >= 1 && j <= N && t == chessboard[i][j] && t == chessboard[i - 1][j + 1] && t == chessboard[i - 2][j + 2] && t == chessboard[i - 3][j + 3] && t == chessboard[i - 4][j + 4])
            return 1;
    }

    return 0;
}