C++第12周项目1——某年的第几天

课程首页在：http://blog.csdn.net/sxhelijian/article/details/11890759

【项目1-当年第几天】
　　定义一个函数，其参数为年、月、日的值，返回这一天为该年的第几天。要求在main函数中输入年月日，然后调用这个函数求值，并在main函数中输出结果。
 

参考解答1：

#include<iostream>
using namespace std;
int days(int y, int m, int d);
int main()
{
    int year, month, day;
    cout<<"输入年 月 日"<<endl;
    cin>>year>>month>>day;
    cout<<"这是该年的第"<<days(year, month, day)<<"天"<<endl;
    return 0;
}

int days(int y, int m, int d)
{
    int sum=d;
    //加上前m-1月的天数
    for(int i=1;i<m;i++)
    {
        switch(i)
        {
        case 2:
            sum+=((y%4==0&&y%100!=0)||y%400==0)?29:28;
            break;
        case 4:
        case 6:
        case 9:
        case 11:
            sum+=30;
            break;
        default:
            sum+=31;
            break;
        }
    }
    return sum;
}

#include<iostream>
using namespace std;
int days(int y, int m, int d);
int main()
{
    int year, month, day;
    cout<<"输入年 月 日"<<endl;
    cin>>year>>month>>day;
    cout<<"这是该年的第"<<days(year, month, day)<<"天"<<endl;
    return 0;
}

int days(int y, int m, int d)
{
    int sum=d;
    //加上前m-1月的天数
    for(int i=1; i<m; i++)
    {
        if(i==1||i==3||i==5||i==7||i==8||i==10||i==12)
            sum+=31;
        else if (i==4||i==6||i==9||i==11)
            sum+=30;
        else
            sum+=((y%4==0&&y%100!=0)||y%400==0)?29:28;
    }
    return sum;
}

下下周学习了数组，可以这样来，30行之内解决问题（对数组充满期待吧）：

#include<iostream>
using namespace std;
int days(int y, int m, int d);
int main()
{
    int year, month, day;
    cout<<"输入年 月 日"<<endl;
    cin>>year>>month>>day;
    cout<<"这是该年的第"<<days(year, month, day)<<"天"<<endl;
    return 0;
}

int days(int y, int m, int d)
{
    int sum=d;
    int a[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
    for(int i=1; i<m; i++)
    {
        sum+=a[i];
    }
    if(m>2&&((y%4==0&&y%100!=0)||y%400==0)) //若闰年，且晚于2月，加一天
        sum++;
    return sum;
}

#include <iostream>
using namespace std;
int shijian(int year,int month,int day);
int main()
{
    int year,month,day,t;
    cout<<"要知道今天是今年第几天吗？请输入年月日："<<endl;
    cin>>year>>month>>day;
    t=shijian(year,month,day);
    cout<<year<<"年"<<month<<"月"<<day<<"是本年第"<<t<<"天"<<endl;
    cout<<"我相信这将是最美好的一天."<<endl;
}
int shijian(int year,int month,int day)
{
    int t;
    if(year%4==0&&year%100!=0||year%400==0)
    {
        switch(month)
        {
        case 1:
            t=day;
            break;
        case 2:
            t=day+31;
            break;
        case 3:
            t=day+31+29;
            break;
        case 4:
            t=day+31+29+31;
            break;
        case 5:
            t=day+31+29+31+30;
            break;
        case 6:
            t=day+31+29+31+30+31;
            break;
        case 7:
            t=day+31+29+31+30+31+30;
            break;
        case 8:
            t=day+31+29+31+30+31+30+31;
            break;
        case 9:
            t=day+31+29+31+30+31+30+31+31;
            break;
        case 10:
            t=day+31+29+31+30+31+30+31+31+30;
            break;
        case 11:
            t=day+31+29+31+30+31+30+31+31+30+31;
            break;
        case 12:
            t=day+31+29+31+30+31+30+31+31+30+31+30;
            break;
        }
        return t;
    }
    else
    {
        switch(month)
        {
        case 1:
            t=day;
            break;
        case 2:
            t=day+31;
            break;
        case 3:
            t=day+31+28;
            break;
        case 4:
            t=day+31+28+31;
            break;
        case 5:
            t=day+31+28+31+30;
            break;
        case 6:
            t=day+31+28+31+30+31;
            break;
        case 7:
            t=day+31+28+31+30+31+30;
            break;
        case 8:
            t=day+31+28+31+30+31+30+31;
            break;
        case 9:
            t=day+31+28+31+30+31+30+31+31;
            break;
        case 10:
            t=day+31+28+31+30+31+30+31+31+30;
            break;
        case 11:
            t=day+31+28+31+30+31+30+31+31+30+31;
            break;
        case 12:
            t=day+31+28+31+30+31+30+31+31+30+31+30;
            break;
        }
        return t;
    }
}