实验2 C语言分支与循环基础应用编程

task1

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

#define N 5

int main()
{
	int number;
	int i;
	
	srand( time(0) );
	
	for(i = 0;i < N; ++i)
	{
		number = rand()%65 + 1;
		printf("20238331%04d\n",number);
	}
	system("pause");
	return 0;
}

问题一：line15的作用是为了控制所输出的数的数值大小在1~65之间；

问题二：该程序是为了提供介于1和65之间的随机数，一共有4位，其中前两位用0替代。

task2

#include<stdio.h>
int main()
{
	char color;
	while((color = getchar())!=EOF)
	{
		if(color == 'r')
		        printf("stop!\n");
		else if(color == 'g')
		        printf("go go go\n");
		else if(color == 'y')
		        printf("wait a minute\n");
		else
		        printf("something must be wrong\n");
		getchar();
		
		    
	}
	return 0;
}

task3

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
int main()
{
    int lucky_day;
	int guess;
	int cnt;
	srand(time(0));
	lucky_day = rand()%31 + 1;
	printf("猜猜2024年哪一天会是你的lucky_day \n开始喽，你有三次机会，猜吧(1~31) : ");    //随机生成lucky_day
	for(cnt = 1;cnt<=3;++cnt)  //猜lucy_day 
	{
		scanf("%d",&guess);
		if(guess < lucky_day)
			printf("你猜的日期早了，你的lucky day还没有到呢\n再猜(1~31): ");
		else if(guess > lucky_day)
			printf("你猜的日期晚了，你的lucky day在前面哦\n再猜(1~31): ");
		else 
		{
			printf("哇，猜中了:-)"); 
			return 0;
		}
		if(cnt == 3)
		    printf("次数用完啦。偷偷告诉你，5月你的lucky_day是%02d号",lucky_day); 
	}
	system("pause");
	return 0; 
}

task4

#include<stdio.h>
#include<stdlib.h>
int main() 
{
	int a,n,d,i;
	float s,j;
	while(scanf("%d%d",&n,&a)!=EOF){
	d = a;
		for(i=1;i<=n;i++){
			j=1.0*i/d;
			s +=j;
			d =d*10+a;
		}
	printf("n = %d , a = %d , s = %f\n",n,a,s);
	}
	return 0;
}
	

task5

#include<stdio.h>
#include<stdlib.h>
int main()
{
	int i,j;
	for(i=1;j<=9;i++)
	{
		for(j=1;j<=i;j++)
		    printf("%d*%d=%2d  ",i,j,i*j);
	    printf("\n");
	}
	return 0;
	    
}

task6

#include<stdio.h>
#include<stdlib.h>
int main()
{   int n,a=0;
	printf("input n: ");
   scanf("%d",&n);
    for(;n>=1;n=n-1)
	{   
	   	for(int m=1;m<=a;m=m+1)
		{ 
			printf("\t");
		}
		for(int i=1;i<=2*n-1;i=i+1)
		{ 
			printf(" O \t");
		}
		printf("\n");
	 	  	for(int m=1;m<=a;m=m+1)
		{ 
			printf("\t");
		}
		for(int j=1;j<=2*n-1;j=j+1)
		{ 
			printf("<H>\t");
		}
		printf("\n");
	  	for(int m=1;m<=a;m=m+1)
		{ 
			printf("\t");
		}
		for(int b=1;b<=2*n-1;b=b+1)
		{ 
			printf("I I\t");
		}
		printf("\n");
		a=a+1;
	
	}
system("pause");
return 0;
}