求本年第几天

/*
*程序的版权和版本声明部分：
*Copyright（c）2013，烟台大学计算机学院学生
*All rights reserved.
*文件名称：
*作者：赵加响
*完成日期：2013年 11月13 日
*版本号：v1.0
*对任务及求解方法的描述部分：
*输入描述:
*问题描述：用函数输出日期是本年的多少天
*程序输出：
*问题分析：
*算法设计：
*/

#include<iostream>
#include<ctime>
using namespace std;
int time(int,int,int);
int main()
{
    int year,month,day,tian;
    cout<<"请输入年月日："<<endl;
    cin>>year>>month>>day;
    tian=time(year,month,day);
    cout<<year<<"年"<<month<<"月"<<day<<"日是本年的第"<<tian<<"天"<<endl;
    return 0;
}
int time(int year,int month,int day)
{
    int tian;
    if(year%4==0&&year%100!=0||year%400==0)
    {
        switch(month)
        {
        case 1:
            tian=day;
            break;
        case 2:
            tian=day+31;
            break;
        case 3:
            tian=day+31+29;
            break;
        case 4:
            tian=day+31+29+31;
            break;
        case 5:
            tian=day+31+29+31+30;
            break;
        case 6:
            tian=day+31+29+31+30+31;
            break;
        case 7:
            tian=day+31+29+31+30+31+30;
            break;
        case 8:
            tian=day+31+29+31+30+31+30+31;
            break;
        case 9:
            tian=day+31+29+31+30+31+30+31+31;
            break;
        case 10:
            tian=day+31+29+31+30+31+30+31+31+30;
            break;
        case 11:
            tian=day+31+29+31+30+31+30+31+31+30+31;
            break;
        case 12:
            tian=day+31+29+31+30+31+30+31+31+30+31+30;
            break;
        }
        return tian;
    }
    else
    {
        switch(month)
        {
        case 1:
            tian=day;
            break;
        case 2:
            tian=day+31;
            break;
        case 3:
            tian=day+31+28;
            break;
        case 4:
            tian=day+31+28+31;
            break;
        case 5:
            tian=day+31+28+31+30;
            break;
        case 6:
            tian=day+31+28+31+30+31;
            break;
        case 7:
            tian=day+31+28+31+30+31+30;
            break;
        case 8:
            tian=day+31+28+31+30+31+30+31;
            break;
        case 9:
            tian=day+31+28+31+30+31+30+31+31;
            break;
        case 10:
            tian=day+31+28+31+30+31+30+31+31+30;
            break;
        case 11:
            tian=day+31+28+31+30+31+30+31+31+30+31;
            break;
        case 12:
            tian=day+31+28+31+30+31+30+31+31+30+31+30;
            break;
        }
        return tian;
    }
}