模拟时间进制转换，并输出当年日历

　　这个是朋友托我打的，不属于ACM，但觉得挺有意思，所以就放到这里了，题目有两个要求，输出当1年1月1日到前年每月的1号隔了多少天，并输出这天是星期几，然后再按格式输出这一年12个月的公历日历。

是一个挺简单的模拟，就是有点小麻烦和小细节需要注意，下面是代码:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
struct Node
{
    int y,m,d,day;
};
Node start;
void init()
{
    start.y = 1;
    start.m = 1;
    start.d = 1;
    start.day = 1;
}
bool is_leap(int year)
{
    if(year%400==0 || (year%4==0&&year%100 != 0))return true;
    else return false;
}
int day_of_month(int year,int m)
{
    if(m==1||m==3||m==5||m==7||m==8||m==10||m==12) return 31;
    if(m==2)
    {
        if(is_leap(year)) return 29;
        return 28;
    }
    return 30;
}
Node tmp;
int getday(Node now)
{
    tmp = start;
    int ans = 0;
    bool flag = false;
    while(tmp.y<=now.y)
    {
        while(tmp.m <= 12)
        {
            while(tmp.d <= day_of_month(tmp.y,tmp.m))
            {
                if(tmp.y==now.y&&tmp.m==now.m&&tmp.d==now.d)
                {
                    flag = true;
                    break;
                }
                tmp.d++;
                tmp.day++;
                if(tmp.day==8)
                {
                    tmp.day = 1;
                }
                ///cout<<"year month day d:"<<tmp.y<<" "<<tmp.m<<" "<<tmp.d<<" "<<tmp.day<<endl;
                ans++;
            }
            if(flag) break;
            tmp.d = 1;
            tmp.m++;
        }
        if(flag) break;
        tmp.m = 1;
        tmp.y++;
    }
    return ans;
}
int main()
{
    int nowyear;
    scanf("%d",&nowyear);
    init();
    Node now;
    now.y = nowyear;
    now.d = 1;
    int firstday[13];
    int allday[13];
    for(int i = 1; i <= 12; i++)
    {
        now.m = i;
        int res = getday(now);
        printf("到%d年%d月1号共有%d天\n",nowyear,i,res);
        printf("这天是星期%d\n",tmp.day);
        firstday[i] = tmp.day;
        allday[i] = day_of_month(nowyear,i);
        puts("");
    }
    for(int i = 1; i <= 12; i++)
    {
        printf("      %d月    \n",i);
        printf("-----------------------\n");
        ///printf("  7  1  2  3  4  5  6\n");
        printf(" 日 一 二 三 四 五 六\n");
        int key = firstday[i];
        ///cout<<"key = "<<key<<endl;
        if(key < 7)
        {
            for(int j = 1;j <= key;j++)
                cout<<"   ";
        }
        int tot = 1,k1 = key==7?7:7-key,k0 = 0;
        bool mark = false;
        while(tot <= allday[i])
        {
            printf("%3d",tot);
            k0++;
            tot++;
            if(k0==k1 && !mark)
            {
                puts("");
                k0 = 0;
                mark = true;
            }
            if(mark&&k0==7)
            {
                puts("");
                k0 = 0;
            }
        }
        puts("");
    }
    return 0;
}