帮我用c语言实现一个俄罗斯方块游戏

好的，下面是一个简单的俄罗斯方块游戏的C语言实现，仅供参考：

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
#include <time.h>
#include <windows.h>

#define BOARD_WIDTH 10      // 游戏区域宽度
#define BOARD_HEIGHT 20     // 游戏区域高度

// 定义俄罗斯方块的形状和颜色
const char block[7][4][4] = {
    { "0000", "1111", "0000", "0000" },
    { "0100", "0100", "0100", "0100" },
    { "0000", "0110", "1100", "0000" },
    { "0010", "0110", "0100", "0000" },
    { "0110", "1100", "0100", "0000" },
    { "0100", "1100", "1000", "0000" },
    { "1000", "1100", "0100", "0000" }
};
const int color[7] = { 1, 2, 3, 4, 5, 6, 7 };

// 定义游戏区域和当前方块的状态
int board[BOARD_HEIGHT + 1][BOARD_WIDTH + 2] = { 0 };
int cur_x, cur_y, cur_block, cur_rotation, cur_color;

// 清屏函数
void clear_screen()
{
    system("cls");
}

// 显示游戏区域和当前方块
void draw_board()
{
    int i, j;
    for (i = 0; i <= BOARD_HEIGHT; i++) {
        for (j = 1; j <= BOARD_WIDTH; j++) {
            printf("%c", board[i][j] ? '#' : ' ');
        }
        printf("\n");
    }
    for (j = 1; j <= 4; j++) {
        for (i = 0; i <= 3; i++) {
            printf("%c", block[cur_block][cur_rotation][i * 4 + j - 1] == '0' ? ' ' : '#');
        }
        printf("     ");
        for (i = 0; i <= 3; i++) {
            printf("%c", board[i + cur_y][j + cur_x] ? '#' : ' ');
        }
        printf("\n");
    }
}

// 检查当前方块是否与已有方块重叠
int check_overlap(int x, int y, int block, int rotation)
{
    int i, j;
    for (i = 0; i <= 3; i++) {
        for (j = 0; j <= 3; j++) {
            if (block[cur_block][rotation][i * 4 + j] != '0' && board[y + i][x + j]) {
                return 1;
            }
        }
    }
    return 0;
}

// 检查当前方块是否到达底部或与已有方块重叠
int check_gameover()
{
    return check_overlap(cur_x, cur_y + 1, cur_block, cur_rotation);
}

// 将当前方块插入到游戏区域中
void insert_block()
{
    int i, j;
    for (i = 0; i <= 3; i++) {
        for (j = 0; j <= 3; j++) {
            if (block[cur_block][cur_rotation][i * 4 + j] != '0') {
                board[cur_y + i][cur_x + j] = cur_color;
            }
        }
    }
}

// 消除已经填满的行
void remove_lines()
{
    int i, j, k, cnt;
    for (i = BOARD_HEIGHT; i >= 1; i--) {
        cnt = 0;
        for (j = 1; j <= BOARD_WIDTH; j++) {
            if (board[i][j]) {
                cnt++;
            }
        }
        if (cnt == BOARD_WIDTH) {
            for (k = i; k > 0; k--) {
                for (j = 1; j <= BOARD_WIDTH; j++) {
                    board[k][j] = board[k - 1][j];
                }
            }
            i++;
        }
    }
}

// 随机生成下一个方块
void new_block()
{
    cur_block = rand() % 7;
    cur_rotation = rand() % 4;
    cur_x = BOARD_WIDTH / 2 - 2;
    cur_y = 0;
    cur_color = color[cur_block];
    if (check_overlap(cur_x, cur_y, cur_block, cur_rotation)) {
        printf("Game Over!\n");
        exit(0);
    }
}

// 处理用户输入
void handle_input()
{
    int key = _getch();
    switch (key) {
        case 'a':
        case 'A':
            if (!check_overlap(cur_x - 1, cur_y, cur_block, cur_rotation)) {
                cur_x--;
            }
            break;
        case 'd':
        case 'D':
            if (!check_overlap(cur_x + 1, cur_y, cur_block, cur_rotation)) {
                cur_x++;
            }
            break;
        case 'w':
        case 'W':
            if (!check_overlap(cur_x, cur_y, cur_block, (cur_rotation + 1) % 4)) {
                cur_rotation = (cur_rotation + 1) % 4;
            }
            break;
        case 's':
        case 'S':
            while (!check_overlap(cur_x, cur_y + 1, cur_block, cur_rotation)) {
                cur_y++;
            }
            break;
    }
}

int main()
{
    srand(time(NULL));
    new_block();
    while (1) {
        clear_screen();
        draw_board();
        Sleep(1000);
        if (check_overlap(cur_x, cur_y + 1, cur_block, cur_rotation)) {
            insert_block();
            remove_lines();
            new_block();
        } else {
            cur_y++;
        }
        if (kbhit()) {
            handle_input();
        }
    }
    return 0;
}

这个游戏实现比较简单，只包含了基本的游戏逻辑和操作，你可以根据自己的需求进行修改和优化。