C语言之每日一练

版权声明：本文为博主原创文章，未经博主允许不得转载。https://blog.csdn.net/qq_42680327   

判断一个数是否为水仙花数：

#include 
void main()
{
    int i,j,k,p;
    printf("Please input number:");
    scanf("%d",&p);
    if(p<99||p>1000){
        printf("%d is not three number!\n",p);
    }
    else{
        i=p/100;
        j=p/10%10;
        k=p%10;
        if(i*i*i+j*j*j+k*k*k==p){
            printf("%d is narcissistic number!\n",p);
        }
        else{
            printf("%d is not narcissistic number!\n",p);
        }
    }
}

打印所有的水仙花数：

#include "stdio.h"
void main()
{
  int i,j,k,m;
  printf("水仙花数分别是:\n");
  for(m=100;m<=999;m++)
  {	
  	i=m/100;        //计算百位
    j=m/10%10;      //计算十位
    k=m%10;         //计算个位
    if(i*100+j*10+k==i*i*i+j*j*j+k*k*k){
        printf("%-5d\n",m);
    }
  }
}

 

从键盘输入一个数，判断输入的数，是否是回文数：

#include 
void main(){
	int num,s,y=0;
	printf("Please input numbers: ");
	scanf("%d", &num);
	s=num;
	while(s>0){
		y=y*10+s%10;
		s=s/10;
	}
	if(y==num){
		printf("%d是一个回文数!\n", num);
	}else{
		printf("%d不是一个回文数!\n", num);
	}
}

闰年判断：

#include 
void main()
{
    int year;
    printf("Please input year:");
    scanf("%d",&year);
    if((year%4==0)||(year%100!=0)&&(year%400==0)){
        printf("%d year is leap year\n",year);
    }else{
         printf("%d year is not leap year\n",year);    
    }
}

从键盘输入三个数a,b,c,判断是否构成三角形：

#include 
void main()
{
    int a,b,c;
    printf("please input trianglea b and c:");
    scanf("%d %d %d",&a,&b,&c);
    if(a+b>c&&a+c>b&&b+c>a){
        printf("%d---%d---%d can form triangle!\n",a,b,c);
    }
    else{
        printf("%d---%d---%d can not form triangle!\n",a,b,c);  
    }
}

从键盘输入年份，月份，判断有多少天（注意闰年）

#include 
void main()
{
    //int a[12]={31,28,31,30,31,30,31,31,30,31,30,31};
    //(year%4==0)||(year%100!=0)&&(year%400==0)
    int year,month,days;
    printf("请输出年份:");
    scanf("%d",&year);
    printf("请输入月份 :");
    scanf("%d",&month);
    switch(month){
    case 1:
    case 3:
    case 5:
    case 7:
    case 8:
    case 10:
    case 12:
        days=31;
        break;
    case 4:
    case 6:
    case 9:
    case 11:
        days=30;
        break;
    case 2:
        if((year%4==0)||(year%100!=0)&&(year%400==0)){
                days=29;
        }
        else{
            days=28;
        }
    }
    printf("%d 年 的 %d 月 有 %d 天!\n",year,month,days);
}