美女程序员200行C语言代码打造贪吃蛇游戏,全场粉丝爆满!

使用C++开发的一款休闲竞技游戏,玩家控制一条小蛇吃掉地图上的食物以增长身体,挑战自己的极限长度。

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今天给大家用C++写一款超好玩的休闲竞技游戏,不仅比拼手速,更考验你的策略!在贪吃蛇大作战的世界中,每个人化身为一条小蛇,通过不断努力变得越来越长,最终成为最贪吃的最长的贪吃蛇!地图以绿地为有点奇怪的贪吃蛇游戏,看上去有些逗,根据vc的MVC编写,源码编译无错,希望大家喜欢。

源码结构截图:

 

主要代码截图:

 

源码测试截图:

 

按键盘上方向键的任何一个键,游戏开始,开始小蛇只有三节,小蛇没吃一个小黑点,蛇身便增长一节,得10分,玩家每次游戏共有6条生命,所有显示都在游戏舞台右侧,方便玩家侧观。

 

 

 

时间函数举例程序分析 2.程序源代码: #include "stdio.h" #include "time.h" void main() { time_t lt; /*define a longint time varible*/ lt=time(NULL);/*system time and date*/ printf(ctime(<)); /*english format output*/ printf(asctime(localtime(<)));/*tranfer to tm*/ printf(asctime(gmtime(<))); /*tranfer to Greenwich time*/ } 【程序92】 题目:时间函数举例2 1.程序分析: 2.程序源代码: /*calculate time*/ #include "time.h" #include "stdio.h" main() { time_t start,end; int i; start=time(NULL); for(i=0;i<3000;i++) { printf("\1\1\1\1\1\1\1\1\1\1\n"); } end=time(NULL); printf("\1: The different is %6.3f\n",difftime(end,start)); } 【程序93】 题目:时间函数举例3 1.程序分析: 2.程序源代码: /*calculate time*/ #include "time.h" #include "stdio.h" main() { clock_t start,end; int i; double var; start=clock(); for(i=0;ii) { printf("please input a little smaller.\n"); scanf("%d",&guess); } else { printf("please input a little bigger.\n"); scanf("%d",&guess); } } end=clock(); b=time(NULL); printf("\1: It took you %6.3f seconds\n",var=(double)(end-start)/18.2); printf("\1: it took you %6.3f seconds\n\n",difftime(b,a)); if(var<15) printf("\1\1 You are very clever! \1\1\n\n"); else if(var<25) printf("\1\1 you are normal! \1\1\n\n"); else printf("\1\1 you are stupid! \1\1\n\n"); printf("\1\1 Congradulations \1\1\n\n"); printf("The number you guess is %d",i); } printf("\ndo you want to try it again?(\"yy\".or.\"n\")\n"); if((c=getch())=='y') goto loop; } 【程序95】 题目:家庭财务管理小程序 1.程序分析: 2.程序源代码: /*money management system*/ #include "stdio.h" #include "dos.h" main() { FILE *fp; struct date d; float sum,chm=0.0; int len,i,j=0; int c; char ch[4]="",ch1[16]="",chtime[12]="",chshop[16],chmoney[8]; pp: clrscr(); sum=0.0; gotoxy(1,1);printf("|----------------------------------------------------|"); gotoxy(1,2);printf("| money management system(C1.0) 2000.03 |"); gotoxy(1,3);printf("|----------------------------------------------------|"); gotoxy(1,4);printf("| -- money records -- | -- today cost list -- |"); gotoxy(1,5);printf("| ------------------------ |-----------------------------|"); gotoxy(1,6);printf("| date: -------------- | |"); gotoxy(1,7);printf("| | | | |"); gotoxy(1,8);printf("| -------------- | |"); gotoxy(1,9);printf("| thgs: ------------------ | |"); gotoxy(1,10);printf("| | | | |"); gotoxy(1,11);printf("| ------------------ | |"); gotoxy(1,12);printf("| cost: ---------- | |"); gotoxy(1,13);printf("| | | | |"); gotoxy(1,14);printf("| ---------- | |"); gotoxy(1,15);printf("| | |"); gotoxy(1,16);printf("| | |"); gotoxy(1,17);printf("| | |"); gotoxy(1,18);printf("| | |"); gotoxy(1,19);printf("| | |"); gotoxy(1,20);printf("| | |"); gotoxy(1,21);printf("| | |"); gotoxy(1,22);printf("| | |"); gotoxy(1,23);printf("|--------------------------------------------------|"); i=0; getdate(&d); sprintf(chtime,"%4d.%02d.%02d",d.da_year,d.da_mon,d.da_day); for(;;) { gotoxy(3,24);printf(" Tab __browse cost list Esc __quit"); gotoxy(13,10);printf(" "); gotoxy(13,13);printf(" "); gotoxy(13,7);printf("%s",chtime); j=18; ch[0 ]=getch(); if(ch[0]==27) break; strcpy (chshop,""); strcpy(chmoney,""); if(ch[0]==9) { mm:i=0; fp=fopen("home.dat","r+"); gotoxy(3,24);printf(" "); gotoxy(6,4);printf(" list records "); gotoxy(1,5);printf("|-------------------------------------|"); gotoxy(41,4);printf(" "); gotoxy(41,5);printf(" |"); while(fscanf(fp,"%10s%14s%f\n",chtime,chshop,&chm)!=EOF) { if(i==36) { getch(); i=0; } if ((i%36)16) { gotoxy(41,4+i-17); printf(" "); gotoxy(42,4+i-17); } i++; sum=sum+chm; printf("%10s %-14s %6.1f\n",chtime,chshop,chm);} gotoxy(1,23);printf("|----------------------------------------------|"); gotoxy(1,24);printf("| |"); gotoxy(1,25);printf("|----------------------------------------------|"); gotoxy(10,24);printf("total is %8.1f$",sum); fclose(fp); gotoxy(49,24);printf("press any key to.....");getch();goto pp; } else { while(ch[0]!='\r') { if(j15) { len=len+1; j=11; } strcpy(ch1,""); j=j-2; strncat(ch1,chtime,len); strcpy(chtime,""); strncat(chtime,ch1,len-1); gotoxy(13,7);printf(" "); } gotoxy(13,7);printf("%s",chtime);ch[0]=getch(); if(ch[0]==9) goto mm; if(ch[0]==27) exit(1); } gotoxy(3,24);printf(" "); gotoxy(13,10); j=0; ch[0]=getch(); while(ch[0]!='\r') { if (j<14) { strncat(chshop,ch,1); j++; } if(ch[0]==8) { len=strlen(chshop)-1; strcpy(ch1,""); j=j-2; strncat(ch1,chshop,len); strcpy(chshop,""); strncat(chshop,ch1,len-1); gotoxy(13,10);printf(" "); } gotoxy(13,10);printf("%s",chshop);ch[0]=getch();} gotoxy(13,13); j=0; ch[0]=getch(); while(ch[0]!='\r') { if (j='a'&&str[i]<='z') str[i]=str[i]-32; fputc(str[i],fp); i++; } fclose(fp); fp=fopen("test","r"); fgets(str,strlen(str)+1,fp); printf("%s\n",str); fclose(fp); } 【程序99】 题目:有两个磁盘文件A和B,各存放一字母,要求把这两个文件中的信息合并(按字母顺序排列), 输出到一个新文件C中。 1.程序分析: 2.程序源代码: #include "stdio.h" main() { FILE *fp; int i,j,n,ni; char c[160],t,ch; if((fp=fopen("A","r"))==NULL) { printf("file A cannot be opened\n"); exit(0); } printf("\n A contents are :\n"); for(i=0;(ch=fgetc(fp))!=EOF;i++) { c[i]=ch; putchar(c[i]); } fclose(fp); ni=i; if((fp=fopen("B","r"))==NULL) { printf("file B cannot be opened\n"); exit(0); } printf("\n B contents are :\n"); for(i=0;(ch=fgetc(fp))!=EOF;i++) { c[i]=ch; putchar(c[i]); } fclose(fp); n=i; for(i=0;i<n;i++) for(j=i+1;jc[j]) { t=c[i];c[i]=c[j];c[j]=t; } printf("\n C file is:\n"); fp=fopen("C","w"); for(i=0;i<n;i++) { putc(c[i],fp); putchar(c[i]); } fclose(fp); } 【程序100】 题目:有五个学生,每个学生有3门课的成绩,从键盘输入以上数据(包括学生号,姓名,三门课成绩),计算出平均成绩,况原有的数据和计算出的平均分数存放在磁盘文件"stud"中。 1.程序分析: 2.程序源代码: #include "stdio.h" struct student { char num[6]; char name[8]; int score[3]; float avr; } stu[5]; main() { int i,j,sum; FILE *fp; /*input*/ for(i=0;i<5;i++) { printf("\n please input No. %d score:\n",i); printf("stuNo:"); scanf("%s",stu[i].num); printf("name:"); scanf("%s",stu[i].name); sum=0; for(j=0;j<3;j++) { printf("score %d.",j+1); scanf("%d",&stu[i].score[j]); sum+=stu[i].score[j]; } stu[i].avr=sum/3.0; } fp=fopen("stud","w"); for(i=0;i<5;i++) if(fwrite(&stu[i],sizeof(struct student),1,fp)!=1) printf("file write error\n"); fclose(fp); }
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