100分求千行C语言程序源代码

Question

希望能有为高手送我一个四千行的C语言程序代码，是能在TC上运行成功的，没有的话就多给几个一千行以上的吧，谢谢大家了！最好快一点，能今天给的追加50分，当然，必须能运行成功的。
在一定程度上来说损失应付吧，但是老师让写七千行，好歹我也辛辛苦苦用一周的时间写了三千行，大家说可以原谅吧

Answer 1

俄罗斯方块，有你的名字：
#include <stdio.h>
#include <dos.h>
#include <conio.h>
#include <graphics.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>

#ifdef __cplusplus
#define __CPPARGS ...
#else
#define __CPPARGS
#endif
#define MINBOXSIZE 23 /* 最小方块的尺寸 */
#define BGCOLOR 7 /* 背景着色 */
#define GX 200
#define GY 10
#define SJNUM 1000 /* 每当玩家打到一千分等级加一级*/

/* 按键码*/
#define VK_LEFT 0x4b00
#define VK_RIGHT 0x4d00
#define VK_DOWN 0x5000
#define VK_UP 0x4800
#define VK_HOME 0x4700
#define VK_END 0x4f00
#define VK_SPACE 0x3920
#define VK_ESC 0x011b
#define VK_ENTER 0x1c0d

/* 定义俄罗斯方块的方向（我定义他为4种）*/
#define F_DONG 0
#define F_NAN 1
#define F_XI 2
#define F_BEI 3

#define NEXTCOL 20 /* 要出的下一个方块的纵坐标*/
#define NEXTROW 12 /* 要出的下一个方块的横从标*/
#define MAXROW 14 /* 游戏屏幕大小*/
#define MAXCOL 20
#define SCCOL -16 /*游戏屏幕大显示器上的相对位置*/
#define SCROW -16

int gril[22][16]; /* 游戏屏幕坐标*/
int col=1,row=7; /* 当前方块的横纵坐标*/
int boxfx=0,boxgs=0; /* 当前寺块的形壮和方向*/
int nextboxfx=0,nextboxgs=0,maxcol=22;/*下一个方块的形壮和方向*/
int minboxcolor=6,nextminboxcolor=6;
int num=0; /*游戏分*/
int dj=0,gamedj[10]={18,16,14,12,10,8,6,4,2,1};/* 游戏等级*/
/* 以下我用了一个3维数组来纪录方块的最初形状和方向*/
int boxstr[7][4][16]={{
{1,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0},
{0,1,0,0,1,1,0,0,1,0,0,0,0,0,0,0},
{1,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0},
{0,1,0,0,1,1,0,0,1,0,0,0,0,0,0,0}},
{
{0,1,1,0,1,1,0,0,0,0,0,0,0,0,0,0},
{1,0,0,0,1,1,0,0,0,1,0,0,0,0,0,0},
{0,1,1,0,1,1,0,0,0,0,0,0,0,0,0,0},
{1,0,0,0,1,1,0,0,0,1,0,0,0,0,0,0}},
{
{1,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0},
{1,1,1,0,1,0,0,0,0,0,0,0,0,0,0,0},
{1,0,0,0,1,0,0,0,1,1,0,0,0,0,0,0},
{0,0,1,0,1,1,1,0,0,0,0,0,0,0,0,0}},
{
{1,1,0,0,1,0,0,0,1,0,0,0,0,0,0,0},
{1,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0},
{0,1,0,0,0,1,0,0,1,1,0,0,0,0,0,0},
{1,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0}},
{
{0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0},
{0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0},
{0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0},
{0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0}},
{
{1,1,0,0,1,1,0,0,0,0,0,0.0,0,0,0},
{1,1,0,0,1,1,0,0,0,0,0,0.0,0,0,0},
{1,1,0,0,1,1,0,0,0,0,0,0.0,0,0,0},
{1,1,0,0,1,1,0,0,0,0,0,0.0,0,0,0}},
{
{0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0},
{1,0,0,0,1,1,0,0,1,0,0,0,0,0,0,0},
{0,1,0,0,1,1,1,0,0,0,0,0.0,0,0,0},
{0,1,0,0,1,1,0,0,0,1,0,0,0,0,0,0}}

};

/* 随机得到当前方块和下一个方块的形状和方向*/
void boxrad()
{
minboxcolor=nextminboxcolor;
boxgs=nextboxgs;
boxfx=nextboxfx;
srand(time(0));
nextminboxcolor=rand()%14+1;
if(nextminboxcolor==1||nextminboxcolor==7||nextminboxcolor==8)
nextminboxcolor=14;
nextboxfx=F_DONG;
srand(time(0));
nextboxgs=rand()%7;
}
/*初始化图形模试*/
void init(int gdrive,int gmode){
int errorcode;
initgraph(&gdrive,&gmode,"..\\bgi");
errorcode=graphresult();
if(errorcode!=grOk){
printf("error of: %s",grapherrormsg(errorcode));
exit(1);
}
}

void f24(int x,int y)
{
static int flag=0;
int i,j,m,k;
int n;
int a[2][66]={{0x00,0x00,0x00,0x00,0x30,0x00,
0x00,0x30,0x00,0x00,0x30,0x00,0x00,0x30,
0x00,0x00,0x30,0xF0,0x04,0x30,0x78,0x0C,
0x30,0x18,0x0C,0x31,0x00,0x0C,0x31,0x80,
0x0C,0x33,0x80,0x0C,0x13,0x00,0x08,0x07,
0x00,0x00,0x06,0x00,0x00,0x0C,0x00,0x00,
0x18,0x00,0x00,0x38,0x00,0x00,0x60,0x00,
0x01,0xC0,0x00,0x03,0x80,0x00,0x0E,0x00,
0x00,0x38,0x00,0x00},
{0x01,0x80,0x00,0x03,0x80,0x00,
0x07,0x06,0x00,0x2E,0x1F,0x00,0x38,0x76,
0x00,0x33,0x66,0x1E,0x13,0x7F,0xFE,0x13,
0x76,0xC6,0x13,0x64,0xC6,0x13,0x7C,0x86,
0x17,0x60,0xBC,0x3F,0x6C,0x9C,0x33,0x6E,
0x8C,0x32,0x7E,0x80,0x06,0xF0,0x80,0x06,
0x60,0x80,0x0C,0x00,0x80,0x0C,0x00,0x80,
0x18,0x00,0x80,0x10,0x00,0x80,0x00,0x00,
0x80,0x00,0x00,0x80} };
flag=(++flag)%2;
for(i=0;i<66;i+=3)
for(k=i;k<i+3;k++)
for(j=0;j<8;j++)
{
n=pow(2,7-j);
m=a[flag][k]/n;
a[flag][k]=a[flag][k]%n;
if(m&&(j+(k-i)*8)!=24&&(j+(k-i)*8)!=23&&(j+(k-i)*8)!=0)
putpixel(x+j+(k-i)*8,y+i/3,1);
}
}

/* 在图形模式下的清屏 */
void cls()
{
setfillstyle(SOLID_FILL,0);
setcolor(0);
bar(0,0,640,480);
}
/*在图形模式下的高级清屏*/
void clscr(int a,int b,int c,int d,int color){
setfillstyle(SOLID_FILL,color);
setcolor(color);
bar(a,b,c,d);
}
/*最小方块的绘制*/
void minbox(int asc,int bsc,int color,int bdcolor){
int a=0,b=0;
a=SCCOL+asc;
b=SCROW+bsc;
clscr(a+1,b+1,a-1+MINBOXSIZE,b-1+MINBOXSIZE,color);
if(color!=BGCOLOR){
f24(a,b+1);
setcolor(bdcolor);
line(a+1,b+1,a-1+MINBOXSIZE,b+1);
line(a+1,b+1,a+1,b-1+MINBOXSIZE);
line(a-1+MINBOXSIZE,b+1,a-1+MINBOXSIZE,b-1+MINBOXSIZE);
line(a+1,b-1+MINBOXSIZE,a-1+MINBOXSIZE,b-1+MINBOXSIZE);
}
}

/*游戏中出现的文字*/
void txt(int a,int b,char *txt,int font,int color){
setcolor(color);
settextstyle(0,0,font);
outtextxy(a,b,txt);
}
/*windows 绘制*/
void win(int a,int b,int c,int d,int bgcolor,int bordercolor){
clscr(a,b,c,d,bgcolor);
setcolor(bordercolor);
line(a,b,c,b);
line(a,b,a,d);
line(a,d,c,d);
line(c,b,c,d);
}

/* 当前方块的绘制*/
void funbox(int a,int b,int color,int bdcolor){
int i,j;
int boxz[4][4];
for(i=0;i<16;i++)
boxz[i/4][i%4]=boxstr[boxgs][boxfx][i];
for(i=0;i<4;i++)
for(j=0;j<4;j++)
if(boxz[i][j]==1)
minbox((j+row+a)*MINBOXSIZE,(i+col+b)*MINBOXSIZE,color,bdcolor);
}
/*下一个方块的绘制 */
void nextfunbox(int a,int b,int color,int bdcolor){
int i,j;
int boxz[4][4];
for(i=0;i<16;i++)
boxz[i/4][i%4]=boxstr[nextboxgs][nextboxfx][i];
clscr((a-1)*MINBOXSIZE,(b-1)*MINBOXSIZE,(3+a)*MINBOXSIZE,(4+b)*MINBOXSIZE,2);
for(i=0;i<4;i++)
for(j=0;j<4;j++)
if(boxz[i][j]==1)
minbox((j+a)*MINBOXSIZE,(i+b)*MINBOXSIZE,color,bdcolor);
}
/*时间中断定义*/
#define TIMER 0x1c
int TimerCounter=0;
void interrupt ( *oldhandler)(__CPPARGS);
void interrupt newhandler(__CPPARGS){
TimerCounter++;
oldhandler();
}
void SetTimer(void interrupt (*IntProc)(__CPPARGS)){
oldhandler=getvect(TIMER);
disable();
setvect(TIMER,IntProc);
enable();
}
/*由于游戏的规则，消掉都有最小方块的一行*/
void delcol(int a){
int i,j;
for(i=a;i>1;i--)
for(j=1;j<15;j++){
minbox(j*MINBOXSIZE,i*MINBOXSIZE,BGCOLOR,BGCOLOR);
gril[i][j]=gril[i-1][j];
if(gril[i][j]==1)
minbox(j*MINBOXSIZE,i*MINBOXSIZE,minboxcolor,0);
}
}

/*消掉所有都有最小方块的行*/
void delete(){
int i,j,zero,delgx=0;
char *nm="00000";
for(i=1;i<21;i++){
zero=0;
for(j=1;j<15;j++)
if(gril[i][j]==0)
zero=1;
if(zero==0){
delcol(i);
delgx++;
}
}
num=num+delgx*delgx*10;
dj=num/10000;

sprintf(nm,"%d",num);
clscr(456,173,500,200,2);
txt(456,173,"Score :",1,4);
txt(456,193,nm,1,4);
}
/*时间中断结束*/
void KillTimer(){
disable();
setvect(TIMER,oldhandler);
enable();
}
/* 测试当前方块是否可以向下落*/
int downok(){
int i,j,k=1,a[4][4];
for(i=0;i<16;i++)
a[i/4][i%4]=boxstr[boxgs][boxfx][i];
for(i=0;i<4;i++)
for(j=0;j<4;j++)
if(a[i][j] && gril[col+i+1][row+j])
k=0;
return(k);
}
/* 测试当前方块是否可以向左行*/
int leftok(){
int i,j,k=1,a[4][4];
for(i=0;i<16;i++)
a[i/4][i%4]=boxstr[boxgs][boxfx][i];
for(i=0;i<4;i++)
for(j=0;j<4;j++)
if(a[i][j] && gril[col+i][row+j-1])
k=0;
return(k);
}
/* 测试当前方块是否可以向右行*/
int rightok(){
int i,j,k=1,a[4][4];
for(i=0;i<16;i++)
a[i/4][i%4]=boxstr[boxgs][boxfx][i];
for(i=0;i<4;i++)
for(j=0;j<4;j++)
if(a[i][j] && gril[col+i][row+j+1])
k=0;
return(k);
}
/* 测试当前方块是否可以变形*/
int upok(){
int i,j,k=1,a[4][4];
for(i=0;i<4;i++)
for(i=0;i<16;i++)
a[i/4][i%4]=boxstr[boxgs][boxfx+1][i];
for(i=3;i>=0;i--)
for(j=3;j>=0;j--)
if(a[i][j] && gril[col+i][row+j])
k=0;
return(k);
}
/*当前方块落下之后，给屏幕坐标作标记*/
void setgril(){
int i,j,a[4][4];
funbox(0,0,minboxcolor,0);
for(i=0;i<16;i++)
a[i/4][i%4]=boxstr[boxgs][boxfx][i];
for(i=0;i<4;i++)
for(j=0;j<4;j++)
if(a[i][j])
gril[col+i][row+j]=1;
col=1;row=7;
}
/*游戏结束*/
void gameover(){
int i,j;

for(i=20;i>0;i--)
for(j=1;j<15;j++)
minbox(j*MINBOXSIZE,i*MINBOXSIZE,2,0);
txt(56,206,"Game Over",4,0);
txt(53,203,"Game Over",4,4);
}
/*按键的设置*/
void call_key(int keyx){
switch(keyx){
case VK_DOWN: { /*下方向键，横坐标加一。*/
if(downok()){
col++;
funbox(0,0,minboxcolor,0);}
else{
funbox(0,0,minboxcolor,0);
setgril();
nextfunbox(NEXTCOL,NEXTROW,4,4);
boxrad();
nextfunbox(NEXTCOL,NEXTROW,nextminboxcolor,0);
delete();
}
break;
}
case VK_UP: { /*上方向键，方向形状旋转90度*/
if(upok())
boxfx++;
if(boxfx>3)
boxfx=0;
funbox(0,0,minboxcolor,0);
break;
}
case VK_LEFT:{ /*左方向键，纵坐标减一*/
if(leftok())
row--;
funbox(0,0,minboxcolor,0);
break;
}
case VK_RIGHT:{ /*右方向键，纵坐标加一*/
if(rightok())
row++;
funbox(0,0,minboxcolor,0);
break;
}
case VK_SPACE: /*空格键，直接落到最后可以落到的们置*/
while(downok())
col++;
funbox(0,0,minboxcolor,0);
setgril();
nextfunbox(NEXTCOL,NEXTROW,4,4);
boxrad();
nextfunbox(NEXTCOL,NEXTROW,nextminboxcolor,0);
delete();
break;
default:
{
txt(423,53,"worng key!",1,4);
txt(428,80,"Plese Enter Anly Key AG!",1,4);
getch();
clscr(420,50,622,97,2);
}
}
}

/*时间中断开始*/
void timezd(void){
int key;
SetTimer(newhandler);
boxrad();
nextfunbox(NEXTCOL,NEXTROW,nextminboxcolor,0);
for(;;){
if(bioskey(1)){
key=bioskey(0);
funbox(0,0,BGCOLOR,BGCOLOR);
if(key==VK_ESC)
break;
call_key(key);
}
if(TimerCounter>gamedj[dj]){
TimerCounter=0;
if(downok()){
funbox(0,0,BGCOLOR,BGCOLOR);
col++;
funbox(0,0,minboxcolor,0);
}
else {
if(col==1){
gameover();
getch();
break;
}
setgril();
delete();
funbox(0,0,minboxcolor,0);
col=1;row=7;
funbox(0,0,BGCOLOR,BGCOLOR);
nextfunbox(NEXTCOL,NEXTROW,4,4);
boxrad();
nextfunbox(NEXTCOL,NEXTROW,nextminboxcolor,0);
}
}
}
}

/*主程序开始*/
void main(void){
int i,j;
char *nm="00000";
init(VGA,VGAHI);
cls();
/*屏幕坐标初始化*/
for(i=0;i<=MAXCOL+1;i++)
for(j=0;j<=MAXROW+1;j++)
gril[i][j]=0;
for(i=0;i<=MAXCOL+1;i++) {
gril[i][0]=1;
gril[i][15]=1;
}
for(j=1;j<=MAXROW;j++){
gril[0][j]=1;
gril[21][j]=1;
}
clscr(0,0,640,480,15);
win(1,1,639,479,2,15);
win(SCCOL+MINBOXSIZE-2,SCROW+MINBOXSIZE-2,SCCOL+15*MINBOXSIZE+2,SCROW+21*MINBOXSIZE+2,BGCOLOR,0);
srand(time(0));
nextboxgs=rand()%7;
nextboxfx=rand()%4;
sprintf(nm,"%d",num);
txt(456,173,"Score :",1,4);
txt(456,193,nm,1,4);
txt(456,243,"Next Box:",1,4);
timezd();
KillTimer();
closegraph();
}

Answer 2

哇，引用一下某人的话，超级怪异的要求，我跟你说一下。可以到gougou.com里面去下。暴多。估计是关键字什么c语言实例。c语言高级之类的。反正多的是。如果我要复制一段代码给你感觉不是很爽。随便复制给你你拿来做什么呢。

Answer 3

送个最简单的给你。
int x=1;
x=2;
x=3;
x=4;
//……
x=7000；

如此类推，七千行。也算应付任务。老师不会怪你的。

Answer 4

恩，4K c代码可以做很多事情了
实在为了应付可以去图书馆找书嘛！

Answer 5

又是应付作业的，没前途。