我来教大家“斗牛怎么押注赢的几率大”(确实是有挂)-哔哩哔哩

网上有关“单片机程序100例”话题很是火热,小编也是针对单片机程序100例寻找了一些与之相关的一些信息进行分析,如果能碰巧解决你现在面临的问题,希望能够帮助到您。

您好:手机麻将有挂是真的吗这款游戏可以开挂,确实是有挂的,咨询加微信【】很多玩家在这款游戏中打牌都会发现很多用户的牌特别好,总是好牌,而且好像能看到其他人的牌一样。所以很多小伙伴就怀疑这款游戏是不是有挂,实际上这款游戏确实是有挂的
http://www.boyicom.net/sheng/1.jpg
1.手机麻将有挂是真的吗这款游戏可以开挂,确实是有挂的,通过添加客服微信 2.咨询软件加微信【】在"设置DD功能DD微信手麻工具"里.点击"开启". 3.打开工具.在"设置DD新消息提醒"里.前两个选项"设置"和"连接软件"均勾选"开启"(好多人就是这一步忘记做了) 4.打开某一个微信组.点击右上角.往下拉."消息免打扰"选项.勾选"关闭"(也就是要把"群消息的提示保持在开启"的状态.这样才能触系统发底层接口)

程序80

题目:海滩上有一堆桃子,五只猴子来分。第一只猴子把这堆桃子凭据分为五份,多了一个,这只猴子把多的一个扔入海中,拿走了一份。第二只猴子把剩下的桃子又平均分成五份,又多了一个,它同样把多的一个扔入海中,拿走了一份,第三、第四、第五只猴子都是这样做的,问海滩上原来最少有多少个桃子?

1.程序分析:

2.程序源代码:

main()

{int i,m,j,k,count;

for(i=4;i<10000;i+=4)

{ count=0;

m=i;

for(k=0;k<5;k++)

{

 j=i/4*5+1;

 i=j;

 if(j%4==0)

count++;

 else

break;

}

 i=m;

 if(count==4)

 {printf("%d\n",count);

break;}

}

}

程序81

题目:809*=800*+9*+1 其中代表的两位数,8*的结果为两位数,9*的结果为3位数。求代表的两位数,及809*后的结果。

1.程序分析:

2.程序源代码:

output(long b,long i)

{ printf("\n%ld/%ld=809*%ld+%ld",b,i,i,b%i);

}

main()

{long int a,b,i;

a=809;

for(i=10;i<100;i++)

{b=i*a+1;

if(b>=1000&&b<=10000&&8*i<100&&9*i>=100)

output(b,i); }

}

==============================================================

程序82

题目:八进制转换为十进制

1.程序分析: 

2.程序源代码:

main()

{ char *p,s[6];int n;

p=s;

gets(p);

n=0;

while(*(p)!='\0')

{n=n*8+*p-'0';

p++;}

printf("%d",n);

}

==============================================================

程序83

题目:求0—7所能组成的奇数个数。

1.程序分析:

2.程序源代码:

main()

{

long sum=4,s=4;

int j;

for(j=2;j<=8;j++)/*j is place of number*/ { printf("\n%ld",sum);

if(j<=2)

s*=7;

else

s*=8;

sum+=s;}

printf("\nsum=%ld",sum);

}

==============================================================

程序84

题目:一个偶数总能表示为两个素数之和。

1.程序分析:

2.程序源代码:

#i nclude "stdio.h"

#i nclude "math.h"

main()

{ int a,b,c,d;

scanf("%d",&a);

for(b=3;b<=a/2;b+=2)

{ for(c=2;c<=sqrt(b);c++)

if(b%c==0) break;

if(c>sqrt(b))

d=a-b;

else

break;

for(c=2;c<=sqrt(d);c++)

if(d%c==0) break;

if(c>sqrt(d))

printf("%d=%d+%d\n",a,b,d);

}

}

==============================================================

程序85

题目:判断一个素数能被几个9整除

1.程序分析:

2.程序源代码:

main()

{ long int m9=9,sum=9;

int zi,n1=1,c9=1;

scanf("%d",&zi);

while(n1!=0)

{ if(!(sum%zi))

n1=0;

else

{m9=m9*10;

sum=sum+m9;

c9++;

}

}

printf("%ld,can be divided by %d \"9\"",sum,c9);

}

==============================================================

程序86

题目:两个字符串连接程序

1.程序分析:

2.程序源代码:

#i nclude "stdio.h"

main()

{char a[]="acegikm";

char b[]="bdfhjlnpq";

char c[80],*p;

int i=0,j=0,k=0;

while(a[i]!='\0'&&b[j]!='\0')

{if (a[i] { c[k]=a[i];i++;}

else

c[k]=b[j++];

k++;

}

c[k]='\0';

if(a[i]=='\0')

p=b+j;

else

p=a+i;

strcat(c,p);

puts(c);

}

==============================================================

程序87

题目:回答结果(结构体变量传递)

1.程序分析: 

2.程序源代码:

#i nclude "stdio.h"

struct student

{ int x;

char c;

} a;

main()

{a.x=3;

a.c='a';

f(a);

printf("%d,%c",a.x,a.c);

}

f(struct student b)

{

b.x=20;

b.c='y';

}

==============================================================

程序88

题目:读取7个数(1—50)的整数值,每读取一个值,程序打印出该值个数的*。

1.程序分析:

2.程序源代码:

main()

{int i,a,n=1;

while(n<=7)

{ do {

 scanf("%d",&a);

 }while(a<1||a>50);

for(i=1;i<=a;i++)

 printf("*");

printf("\n");

n++;}

getch();

}

==============================================================

程序89

题目:某个公司采用公用电话传递数据,数据是四位的整数,在传递过程中是加密的,加密规则如下:每位数字都加上5,然后用和除以10的余数代替该数字,再将第一位和第四位交换,第二位和第三位交换。

1.程序分析:

2.程序源代码:

main()

{int a,i,aa[4],t;

scanf("%d",&a);

aa[0]=a%10;

aa[1]=a%100/10;

aa[2]=a%1000/100;

aa[3]=a/1000;

for(i=0;i<=3;i++)

 {aa[i]+=5;

 aa[i]%=10;

 }

for(i=0;i<=3/2;i++)

 {t=aa[i];

 aa[i]=aa[3-i];

 aa[3-i]=t;

 }

for(i=3;i>=0;i--)

printf("%d",aa[i]);

}

==============================================================

程序90

题目:专升本一题,读结果。

1.程序分析:

2.程序源代码:

#i nclude "stdio.h"

#define M 5

main()

{int a[M]={1,2,3,4,5};

int i,j,t;

i=0;j=M-1;

while(i {t=*(a+i);

*(a+i)=*(a+j);

*(a+j)=t;

i++;j--;

}

for(i=0;i printf("%d",*(a+i));

}

程序91

题目:时间函数举例1

1.程序分析:

2.程序源代码:

#i nclude "stdio.h"

#i nclude "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*/ #i nclude "time.h"

#i nclude "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*/ #i nclude "time.h"

#i nclude "stdio.h"

main()

{ clock_t start,end;

int i;

double var;

start=clock();

for(i=0;i<10000;i++)

{ printf("\1\1\1\1\1\1\1\1\1\1\n");}

end=clock();

printf("\1: The different is %6.3f\n",(double)(end-start));

}

==============================================================

程序94

题目:时间函数举例4,一个猜数游戏,判断一个人反应快慢。

1.程序分析:

2.程序源代码:

#i nclude "time.h"

#i nclude "stdlib.h"

#i nclude "stdio.h"

main()

{char c;

clock_t start,end;

time_t a,b;

double var;

int i,guess;

srand(time(NULL));

printf("do you want to play it.('y' or 'n') \n");

loop:

while((c=getchar())=='y')

{

i=rand()%100;

printf("\nplease input number you guess:\n");

start=clock();

a=time(NULL);

scanf("%d",&guess);

while(guess!=i)

{if(guess>i)

{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*/ #i nclude "stdio.h"

#i nclude "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)<17)

{ gotoxy(4,6+i);

printf(" ");

gotoxy(4,6+i);}

else

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(j<10)

{ strncat(chtime,ch,1);

j++;}

if(ch[0]==8)

{

len=strlen(chtime)-1;

if(j>15)

{ 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<6)

{ strncat(chmoney,ch,1);

j++;}

if(ch[0]==8)

{ len=strlen(chmoney)-1;

strcpy(ch1,"");

j=j-2;

strncat(ch1,chmoney,len);

strcpy(chmoney,"");

strncat(chmoney,ch1,len-1);

gotoxy(13,13);printf(" ");}

gotoxy(13,13);printf("%s",chmoney);ch[0]=getch();}

if((strlen(chshop)==0)||(strlen(chmoney)==0))

continue;

if((fp=fopen("home.dat","a+"))!=NULL);

fprintf(fp,"%10s%14s%6s",chtime,chshop,chmoney);

fputc('\n',fp);

fclose(fp);

i++;

gotoxy(41,5+i);

printf("%10s %-14s %-6s",chtime,chshop,chmoney);

}}}

==============================================================

程序96

题目:计算字符串中子串出现的次数

1.程序分析:

2.程序源代码:

#i nclude "string.h"

#i nclude "stdio.h"

main()

{ char str1[20],str2[20],*p1,*p2;

int sum=0;

printf("please input two strings\n");

scanf("%s%s",str1,str2);

p1=str1;p2=str2;

while(*p1!='\0')

{

if(*p1==*p2)

{while(*p1==*p2&&*p2!='\0')

{p1++;

p2++;}

}

else

p1++;

if(*p2=='\0')

sum++;

p2=str2;

}

printf("%d",sum);

getch();}

==============================================================

程序97

题目:从键盘输入一些字符,逐个把它们送到磁盘上去,直到输入一个#为止。

1.程序分析: 

2.程序源代码:

#i nclude "stdio.h"

main()

{ FILE *fp;

char ch,filename[10];

scanf("%s",filename);

if((fp=fopen(filename,"w"))==NULL)

{printf("cannot open file\n");

exit(0);}

ch=getchar();

ch=getchar();

while(ch!='#')

{fputc(ch,fp);putchar(ch);

ch=getchar();

}

fclose(fp);

}

==============================================================

程序98

题目:从键盘输入一个字符串,将小写字母全部转换成大写字母,然后输出到一个磁盘文件“test”中保存。输入的字符串以!结束。

1.程序分析:

2.程序源代码:

#i nclude "stdio.h"

main()

{FILE *fp;

char str[100],filename[10];

int i=0;

if((fp=fopen("test","w"))==NULL)

{ printf("cannot open the file\n");

exit(0);}

printf("please input a string:\n");

gets(str);

while(str[i]!='!')

{ if(str[i]>='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,各存放一行字母,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:

#i nclude "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;j<n;j++)

if(c[i]>c[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:

#i nclude "stdio.h"

struct student

{ char num[6];

char name[8];

int score[3];

float avr;

} stu[5];

main()

{int i,j,sum

两只数码管段码分别由P0,P2输出控制,P0是十位,P2是个位。P3.2,P3.4分别接两只按键,P32.清0,P3.4计数。

以上程序来自《单片机C语言程序设计实训100例——基于8051+Proteus仿真》

#include <reg52.h>

#define uchar unsigned char

#define uint unsigned int

uchar DSY_CODE[]=

{

0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f

};

uchar Count = 0;

void main()

{

P0 = 0x00;

P2 = 0x00;

TMOD = 0x06;

TH0=255;

TL0=255;

ET0=1;

EX0=1;

EA =1;

IP =0x02;

IT0=1;

TR0=1;

while(1)

{

P0 = DSY_CODE[Count/10];

P2 = DSY_CODE[Count%10];

}

}

void Clear_Counter() interrupt 0

{

Count = 0;

}

void Key_Counter() interrupt 1

{

Count = (Count + 1) %100;

}

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