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;    }}