Basic exercises of C language

Topic 1 ： Yes 1、2、3、4 A digital , How many different and unrepeated three digit numbers can be formed ？ How many are they ？

#include<stdio.h>
 
int main()
{
    int i,j,k,l=0;
	

    printf(" subject ： Yes 1、2、3、4 A digital , How many different and unrepeated three digit numbers can be formed ？ How many are they ？\n");
    for(i=1;i<5;i++) { //  Here's a triple cycle 
		 for(j=1;j<5;j++) {
			 for(k=1;k<5;k++){	 
				 if(i!=k&&k!=j&&i!=j){
				 printf("%d,%d,%d\n",i,j,k);
				l+=1;
				 }
			 }
        	
		 }
    }
	printf(" Altogether %d That's ok ",l);
}

Topic two ： seek 100-999 All the daffodils in the book , So-called “ Narcissistic number ” A three digit number , The sum of its cubes is equal to the number itself .

#include <stdio.h>
int main(){
	int a,i,j,k;
	for(a=100;a<1000;a++){
		i = a/100%10;
		j = a/10%10;
		k = a%10;
		if(k*k*k+j*j*j+i*i*i==a){
			printf("%d\n",a);
		}
	}
}

Topic three ： Enter the date of the year , Judge how many days have passed in this year by this day ？

#include <stdio.h>

int main()
{
	int year,month,day,sum;
       printf(" Please enter the date , The format is 2022,3,28\n");
	scanf("%d,%d,%d",&year,&month,&day);
	switch(month){
		case 1:sum=0;break;
		case 2:sum=31;break;
		case 3:sum=59;break;
		case 4:sum=90;break;
		case 5:sum=120;break;
		case 6:sum=151;break;
		case 7:sum=181;break;
		case 8:sum=212;break;
		case 9:sum=243;break;
		case 10:sum=273;break;
		case 11:sum=304;break;
		case 12:sum=334;break;
		default:printf(" Month error ");break;
	}
	 sum=sum+day;
	/* Detect whether it is a leap month */
	if(year%400==0 || year%4==0 && year%100!=0){
		if(month > 2){
			sum = sum + 1;
		}
	}
   printf("%d Year ended %d month %d The day passed altogether %d God ",year,month,day,sum);
   return 0;
}

Topic four ： Programming calculation 100 to 999 How many numbers between , The sum of the figures is 5 .

#include <stdio.h>
int main(){
	/*i For hundreds ,j by 10 position ,k For a bit */
	int i,j,k,num;
	for (num=100;num<999;num++){
		i = num/100;
		j = num /10 % 10;
		k = num % 10;
		if(i+j+k == 5){
			printf("%d\n",num);
		}
	}
	printf(" end ");
	return 0; 
}

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