C语言如果我要写一个小程序 任意输入两个整数 求出他们之间的所有整数 该怎么写啊 求教

{ printf("Hello, how are you?");

int x;

c语言小程序_c语言小程序编程

c语言小程序_c语言小程序编程

}void move1()//注意只是循环输出各个数字，不能对循环输出再进行循环（如果对循环输出0123456789再进行循环，则move1就变成一个无限循环的函数，则下面的screen循环就进行不下去了）

scanf("%d",&x);

scanf("%d",&y);

int v=x>y?-1:1;

for(;x!=y;x+=v)

printf("%比如输入：dn",x);

printf("%dn",y);

C语言编写小程序

5.printf 形参逗号

#include

int main()

{float a[5][4],er[5]={0},sum[5]={0},high=0.0;

char name[5][20];

int i,j,k;

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

{printf("请输入第%d位学生的名字:",i+1);

scanf("%s",name[i]);

printf("请输入第%d位学生的成绩:",i+1);

for ( j = 0 ; j < 4 ; j++ )

{scanf("%f",&a[i][j]);

sum[i] += a[i][j];

}if ( sum[i] > high)

{high = sum[i];

k = i ;

}}

for ( i = 0 ; i < 4 ; i++ )

{for ( j =0 ; j < 4 ; j++ )

}er[i] /= 4;

}prin{int x,i;tf("n");

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

printf("%s的总分是:%.2fn",name[i],sum[i]);

printf("n");

for ( i = 0 ; i < 4 ; i++ )

print#include f("第%d门课的平均分是:%.2fn",i+1,er[i]);

printf("nn分是:%s,他的总分是:%.2fn",name[k],sum[k]);

printf("n");

c语言小程序，求结果和解答。

（注意：这不是54main() //C语言中的主函数，C程序都从这里开始执行32（五千四百三十二）而是由（5，4，3，2）4个数字组成的）

是编译器的原因，float一般是八字节，int是二或四字节，将int类型强制转换float之后是向前扩展（补码）。但是void swap(char a,char b)你是输出int型，所以前面的都是0000……。你将数据设置成负数，效果就很明显了！

欢迎采纳！

C语言的写一个100元循环购买东西的小程序，谢谢大家了

}

typedef struct goods

int pr;

}GOODS_T;

int main(int argc, char argv[])

{int money_all = 0;

int money_now = 0;

int i = 0;

GOODS_T goods[MAX_GOODS] = {0};

printf("请输入预算：");

scanf("%d", &money_all);

while(i < MAX_GOODS)

{printf("请输入项目和价格：");

scanf("%d %d", &goods[i]t, &goods[i{p=m;].pr);

money_now += (goods[i]t goods[i].pr);

if (money_now == money_all)

{printf("您的剩余预算是0。

");

break;

}else if (money_now > money_all)

{money_now -= (goods[i]t goods[i].pr);

printf("你不能买的东西。

");

printf("您的剩余预算是%d。

", money_all - money_now);

break;

}igoto start;++;

}return 0;

}要注意的是，这份代码里输入项目和价格时用空格来区分数量和单价，不能用标点符号，如：请输入项目和价格：10 4

c语言编个小程序

("pause");

这个明显是C++写的 用到了C++的输入输出流的写法 还有那个iostream.h的头文件 hoho ~~~~

{int j,t;}

cout<<"输入若干个字符串"<

地方官

c语言中怎样将万年历与计时俩程序整和在一起

void main()

/C语言编写万年历问题

16 }

这篇文章主要介绍了C语言实现的一个万年历小程序,void move2()相对简单,做C语言日期计算的朋友可以参考下

该程序简单地输入一个年份（1901年之后的年份），随后程序输出该年份十二个月的日历。/#define Mon 1 #define Tues 2 #define Wed 3 #define Thur 4 #define Fri 5 #define Sat 6 #define Sun 0 #define January_days 31 #define February_days 28 #define March_days 31 #define April_days 30 #define May_days 31 #define June_days 30 #define July_days 31 #define August_days 31 #define September_days 30 #define October_days 31 #define November_days 30 #define December_days 31 #define first1month January_days #define first2month January_days+February_days #define first3month January_days+February_days+March_days #define first4month January_days+February_days+March_days+April_days #define first5month January_days+February_days+March_days+April_days+May_days #define first6month January_days+February_days+March_days+April_days+May_days+June_days #define first7month January_days+February_days+March_days+April_days+May_days+June_days +July_days #define first8month January_days+February_days+March_days+April_days+May_days+June_days +July_days+August_days #define first9month January_days+February_days+March_days+April_days+May_days+June_days +July_days+August_days+September_days #define first10month January_days+February_days+March_days+April_days+May_days+June_days +July_days+August_days+September_days+October_days #define first11month January_days+February_days+March_days+April_days+May_days+June_days +July_days+August_days+September_days+October_days+November_days int known_weekday = Tues; int known_year = 1901; int konwn_month = 1; int known_day = 1; int day_count(int month) { switch(month) { case 1: return 0;break; case 2: return first1month;break; case 3: return first2month;break; case 4: return first3month;break; case 5: return first4month;break; case 6: return first5month;break; case 7: return first6month;break; case 8: return first7month;break; case 9: return first8month;break; case 10: return first9month;break; case 11: return first10month;break; case 12: return first11month;break; } } /char month_name(int month) { switch(month) { case 1: return "January"; break; case 2: return "February"; break; case 3: return "March"; break; case 4: return "April"; break; case 5: return "May"; break; case 6: return "June"; break; case 7: return "July"; break; case 8: return "August"; break; case 9: return "September"; break; case 10: return "October"; break; case 11: return "November"; break; case 12: return "December"; break; default: break; } }/ char month_name(int month) { switch(month) { case 1: return "一月"; break; case 2: return "二月"; break; case 3: return "三月"; break; case 4: return "四月"; break; case 5: return "五月"; break; case 6: return "六月"; break; case 7: return "七月"; break; case 8: return "八月"; break; case 9: return "九月"; break; case 10: return "十月"; break; case 11: return "十一月"; break; case 12: return "十二月"; break; default: break; } } /按格式打印某年某月名称/void first_line_print(int month, int year) { //printf("%20s %d n",month_name(month),year); printf("%8d年 %s n",year, month_name(month)); } /按格式打印星期名称/void week_print() { //printf("%-6s%-6s%-6s%-6s%-6s%-6s%-6sn","Sun.", "Mon.", "Tues.", "Wed.", "Thur.", "Fri.", "Sat."); printf("%-3s%-3s%-3s%-3s%-3s%-3s%-3sn","日", "一", "二", "三", "四", "五", "六"); } /计算该年该月份与已知日子之间的距离天数/int date_distance_count(int month, int year) { int leap_year_count = 0; int i; int distance; if (year > known_year) { for (i=known_year; i 2) { if(((year%4 == 0) && (year%100 != 0) ) || (year%400 == 0)) { leap_year_count++; } } } else if (year == known_year) { if (month > 2) { leap_year_count = 1; } } distance = (year - known_year)365 + leap_year_count + day_count(month); return distance; } /确定该月份天是星期几/int makesure_firstday_weekday(int month, int year) { int date_distance = 0; int weekday; date_distance = date_distance_count(month, year); weekday = (known_weekday + date_distance)%7; return weekday; } /依次打印出该月份的日子/void print_in_turn(int month, int firstday, int year) { int i = 1; int weekday; switch(firstday) { case Sun: break; case Mon: printf("%-3s",""); break; case Tues: printf("%-6s",""); break; case Wed: printf("%-9s",""); break; case Thur: printf("%-12s",""); break; case Fri: printf("%-15s",""); break; case Sat: printf("%-18s",""); break; } switch(month) { case 1: case 3: case 5: case 7: case 8: case 10: case 12: { for(i=0; i<31; i++) { weekday = (firstday + i)%7; printf("%-3d",i+1); if(weekday == Sat) { printf("n"); } } break; } case 2: { if(((year%4 == 0) && (year%100 != 0) ) || (year%400 == 0)) /闰年/ { for(i=0; i<29; i++) { weekday = (firstday + i)%7; printf("%-3d",i+1); if(weekday == Sat) { printf("n"); } } break; } else /平年/ { for(i=0; i<28; i++) { weekday = (firstday + i)%7; printf("%-3d",i+1); if(weekday == Sat) { printf("n"); } } break; } } case 4: case 6: case 9: case 11: { for(i=0; i<30; i++) { weekday = (firstday + i)%7;/计算该天是星期几/ printf("%-3d",i+1); if(weekday == Sat) { printf("n");/如果是星期六，则换行/ } } break; } } } void date_print(int month, int year) { int firstday; firstday = makesure_firstday_weekday(month, year); print_in_turn(month, firstday, year); printf("n"); } void main_month(int month, int year) { first_line_print(month, year); week_print(); date_print(month, year); printf("nn"); } void main_calendar(int year) { int i; for(i=1; i<=12; i++) { main_month(i, year); } } int main() { int year; printf("请输入年份：year = "); scanf("%d",&year); printf("n"); while(year < 1902) { printf("请输入大于1901的年份n"); printf("请输入年份：year = "); scanf("%d",&year); printf("n"); } main_calendar(year); scanf(" "); return 0; }

用C语言编写一个小程序，要求从1+2+3+……+10000，然后把结果输出.

用递归调用写了一个小程序：

int fun (int n)

{int a=0;

if(n==1)

a=1;

else a=n+fun(n-1);

return a;

}void main()

{int fun(int n);

int n;

scanf("%d",&n);

printf("输入的数为%dn",n);

y=fun(n);

你将他们回答中的变量类型变为长整型（long），好像结果要超出int范围

#include

for(int i=1;i<10001;i++)

sum+=cin>>a;num;

printf("结果是%d",sum);

{int i;

printf("i = ");

scanint y;f("%d",&i);

printf("Add 1 to %d = %d",i,Add(i));

}long Add(int i)

{return (((i+1)i)>>1);

#include

main()

for(a=1;a<=10000;a++)

{sum=sum+a;

}print如：输入123，打出321f ("sum=%d",sum);

#include

for(num=1;num<=10000;num++)

sum+=num;

printf("结果是%d",sum);

这是一个C语言小程序，请问它是什么意思

#include //编译时加入的标准I/scanf("%d",&num);O头文件

{char ch1='A',ch2='a'; //定义ch#include1,ch2为字符型的变量，并赋值'A'和'a'

printf("%cn",(ch1,chscanf("%ld",&number);2)); //在屏幕上显示ch1和ch2的值后换行

C语言小程序解疑

if(m%4c

那个4是宽度，也就是8 if(dics<0);说你输入不够4个字符时scanf是不会读取键盘缓存区的。当你输入够了他就读但还是只用一个字符。

abcd7890

a 7

我们可以看出scanf将4个一组取回把再把个赋给c1、c2

顺便说一下：你的iostream忘加.h了

个%4c对应c1.第二个%4c对应c2.

c前面的数字4是限制域宽

也就是限制c1.c2所占的空间

加一个4是指输出的位数一共4位,占4个字符

如果输出a,那么输出的是 a,在a前面有3个空格

为好看~整齐用for(i=1;i<=10;i++)的~不管他

李示羊 - 五级 3-27 18:19

回答的才叫正确～

求一个用C语言编写的小游戏代码

/也不知道你是什么级别的，我是一个新手，刚接触编程语言，以下是我自己变得一个小程序，在所有c语言的编译器（vc++6.0、turbo…………）上都能运行，你还可以进一步改进。这是一个类似贪吃蛇的小游戏。祝你好运/

/贪吃蛇/

#include

#include

#include

int head=3 ,tail=0;

int main()

{int i,j,k=0;

int zuobiao[2][80];

long start;

int direction=77;

int speed;

int timeover;

int change(char qipan[20][80],int zuobiao[2][80],char direction);

zuobiao[0][tail]=1;zuobiao[1][tail]=1;zuobiao[0][1]=1;zuobiao[1][1]=2;zuobiao[0][2]=1;zuobiao[1][2]=3;zuobiao[0][head]=1;zuobiao[1][head]=4;

/处理棋盘/

char qipan[20][80];//定义棋盘

for(i=0;i#define MAX_GOODS 100<20;i++)

for(j=0;j<80;j++)

qipan[i][j]=' ';//初始化棋盘

for(i=0;i<80;i++)

qipan[0][i]='_';

for(i=0;i<20;i++)

qipan[i][0]='|';

for(i=0;i<20;i++)

qipan[i][79]='|';

for(i=0;i<80;i++)

qipan[19][i]='_';

qipan[1][1]=qipan[1][2]=qipan[1][3]='';//初始化蛇的位置

qipan[1][4]='#';

printf("This is a of a SNAKE.nGOOD LUCK TO YOU !n");

printf("Input your speed,please.(e.g.300)n");

scanf("%d",&speed);

while(direction!='q')

{("cls");

for(j=0;j<80;j++)

printf("%c",qipan[i][j]);

timeover=1;

start=clock();

while(!kbhit()&&(timeover=clock()-start<=speed));

if(timeover)

{getch();

direction=getch();

direction=direction;

if(!(direction==72||direction==80||direction==75||direction==77))

{return 0;

("cls");

printf("GAME OVER!n");

}if(!change(qipan,zuobiao,direction))

{direction='q';

("cls");

printf("GAME OVER!n");

}}

}int change(char qipan[20][80],int zuobiao[2][80],char direction)

{int x,y;

if(direction==72)

x=zuobiao[0][head]-1;y=zuobiao[1][head];

if(direction==80)

x=zuobiao[0][head]+1;y=zuobiao[1][head];

if(direction==75)

x=zuobiao[0][head];y=zuobiao[0][head]-1;

if(direction==77)

x=zuobiao[0][head];y=zuobiao[1][head]+1;

if(x==0||x==18||y==78||y==0)

if(qipan[x][y]!=' ')

qipan[zuobiao[0][tail]][zuobiao[1][tail]]=' ';

tail=(tail+1)%80;

qipan[zuobiao[0][head]][zuobiao[1][head]]='';

head=(head+1)%80;

zuobiao[0][head]=x;

zuobiao[1][head]=y;

qipan[zuobiao[0][head]][zuobiao[1][head]]='#';

return 1;

我也是一个新手。

下面有一个游戏

名字叫：

“坑人的无限”（一）：

using namespace std;

int a;

class Screen

{private:

int n;

public:

Screen()

{n=5;

{for(int i=0;i<10;++i)

{cout<

}}

{char i;

for(i='a';i<='z';++i)

{cout<

}}

void screen()

{int t;

while(!kbhit())

{t=time(0)%(2n);//如果是放在循环外面的话，time（0）的值就一直不变，放在循环里面，一秒钟进行一次判断，一秒钟进行一次循环

if(t

move1();

else

move2();

}}

};

int main(){

cout<<"欢迎来到“无限 ”游戏"<

cout<<"下面会输出无限个笑脸"<

cout<<"按'enter'取消"<

Sleep(4000);

Screen s;

s.screen();

cout<

cout<<"接下来会更让你丧心病狂的!"<

cout<<"但是坚持过后必有彩蛋！！！！！！加油！！";

cout<

Sleep(10000);

for(int as=0;as<=50;as++){

for(int i=0;i<=100;i++){

for(int j=0;j<=10;j++){

cout<

}cou1.#include "stdio.h" ==> #include t<

}for(int i=0;i<=100;i++){

for(int j=0;j<=10;j++){

cout<

}cout<

}}

cout<<"你居然坚持下来了！"<

for(int i=0;i<=10;i++){

cout<<"-----------------------------------------------"<

}cout<<"敬请期待！等待无限游戏（二）！";