C语言之生肖、年龄、星座查询

程序实现输入年月日

输出星座生肖年龄

代码实现:

#include<stdio.h>
#include<time.h>
#include<stdlib.h>
int main()
{
	struct tm *Tm;
	time_t t;
	time(&t);
	Tm = localtime(&t);
	
	int toyear = Tm->tm_year+1900;
	int tomon = Tm->tm_mon+1;
	int today = Tm->tm_mday;
	int year,mon,day;	
    char * xz[12]={"摩羯座","水瓶座","双鱼座","白羊座","金牛座","双子座","巨蟹座","狮子座","处女座","天平座","天蝎座","射手座"};
    printf("请输入您的出生年月日(示例:1999 4 23):");
	scanf("%d%d%d",&year,&mon,&day);    
	int m = mon;
	int n = day;
    int shengxiao = 0;
	//星座判断 
        switch (m ) {
            case 1: printf("您的星座是%s\n",n<=19?xz[0]:xz[1]);
                break;
            case 2: printf("您的星座是:%s\n",n<=18?xz[1]:xz[2]);
                break;
            case 3: printf("您的星座是:%s\n",n<=20?xz[2]:xz[3]);
                break;
            case 4: printf("您的星座是%s\n",n<=19?xz[3]:xz[4]);
                break;
            case 5: printf("您的星座是:%s\n",n<=20?xz[4]:xz[5]);
                break;
            case 6: printf("您的星座是:%s\n",n<=21?xz[5]:xz[6]);
                break;
            case 7: printf("您的星座是:%s\n",n<=22?xz[6]:xz[7]);
                break;
            case 8: printf("您的星座是:%s\n",n<=22?xz[7]:xz[8]);
                break;
            case 9: printf("您的星座是:%s\n",n<=22?xz[8]:xz[9]);
                break;
            case 10:printf("您的星座是:%s\n",n<=23?xz[9]:xz[10]);
                break;
            case 11:printf("您的星座是:%s\n",n<=22?xz[10]:xz[11]);
                break;
            case 12:printf("您的星座是:%s\n",n<=21?xz[11]:xz[0]);
                break;
            default:printf("ERROR!!!!");
                break;
        }
        //从1900年开始生肖为鼠 
    	shengxiao = (year - 1899) %12;
        switch (shengxiao) {
            case 1: printf("您的生肖是:鼠\n");
                break;
            case 2: printf("您的生肖是:牛\n");
                break;
            case 3: printf("您的生肖是:虎\n");
                break;
            case 4: printf("您的生肖是:兔\n");
                break;
            case 5: printf("您的生肖是:龙\n");
                break;
            case 6: printf("您的生肖是:蛇\n");
                break;
            case 7: printf("您的生肖是:马\n");
                break;
            case 8: printf("您的生肖是:羊\n");
                break;
            case 9: printf("您的生肖是:猴\n");
                break;
            case 10:printf("您的生肖是:鸡\n");
                break;
            case 11:printf("您的生肖是:狗\n");
                break;
            case 12:printf("您的生肖是:猪\n"); 
                break;
            default:printf("ERROR!!!!");
                break;
        }
        //年龄计算
		if(mon<tomon) 
		printf("您现在的年龄为:%d",toyear-year);
		else if(mon>tomon) 
		printf("您现在的年龄为:%d",toyear-year-1);
		else if(mon<tomon) 
		{
			if(day<today)
			printf("您现在的年龄为:%d",toyear-year);
			else
			printf("您现在的年龄为:%d",toyear-year-1);		
		}

		 
	return 0;
}

程序难点在于当前时间的获取:

struct tm *Tm;
time_t t;
time(&t);
Tm = localtime(&t);

int toyear = Tm->tm_year+1900;
int tomon = Tm->tm_mon+1;
int today = Tm->tm_mday;

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转载自blog.csdn.net/qq_43581670/article/details/107605493