ccf 日期问题求答案求答案

问题描述
　　给定一个年份y和一个整数d，问这一年的第d天是几月几日？
　　注意闰年的2月有29天。满足下面条件之一的是闰年：
　　1）年份是4的整数倍，而且不是100的整数倍；
　　2）年份是400的整数倍。
输入格式
　　输入的第一行包含一个整数y，表示年份，年份在1900到2015之间（包含1900和2015）。
　　输入的第二行包含一个整数d，d在1至365之间。
输出格式
　　输出两行，每行一个整数，分别表示答案的月份和日期。
样例输入
2015
80
样例输出
3
21
样例输入
2000
40
样例输出
2
9

 #pragma once
class CDate
{

public:
    CDate();
    CDate(unsigned year, unsigned month, unsigned day) :m_nYear(year), m_nMonth(month), m_nDay(day){}
    virtual ~CDate();

public:
    friend int DaysMinus(const CDate& smallDate, const CDate& bigDate);
private:
    unsigned int m_nYear;
    unsigned int m_nMonth;
    unsigned int m_nDay;

};

#include "stdafx.h"
#include "CDate.h"


CDate::CDate()
{
    m_nYear = 0;
    m_nMonth = 0;
    m_nDay = 0;
}


CDate::~CDate()
{
}

int DaysMinus(const CDate& smallDate, const CDate& bigDate)
{
    if (smallDate.m_nYear > bigDate.m_nYear)
        cout << "smallDate is bigger than bigDate ！"<<endl;
    else if(smallDate.m_nYear==bigDate.m_nYear){
        if (smallDate.m_nMonth>bigDate.m_nMonth)
            cout << "smallDate is bigger than bigDate ！"<<endl;
        else if (smallDate.m_nMonth==bigDate.m_nYear&&smallDate.m_nDay>bigDate.m_nDay)
            cout << "smallDate is bigger than bigDate ！" << endl;
    }

    int days = 0;//相差的天数
    for (int i = smallDate.m_nYear + 1; i < bigDate.m_nYear; i++){
        days += 365;
        if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0)//闰年的条件
            days++;
    }

    int iSmdays = 0,iBidays = 0;//第一年和最后一年的时间
    bool bSm_Run, bBi_Run;

    bSm_Run = ((smallDate.m_nYear % 4 == 0 && smallDate.m_nYear % 100 != 0) || smallDate.m_nYear % 400 == 0);//小年是闰年

    switch (smallDate.m_nMonth)
    {
    case 1:iSmdays = smallDate.m_nDay; break;
    case 2:iSmdays = 31 + smallDate.m_nDay; break;
    case 3:iSmdays = 31 + 28 + smallDate.m_nDay + bSm_Run; break;
    case 4:iSmdays = 31 * 2 + 28 + smallDate.m_nDay + bSm_Run; break;
    case 5:iSmdays = 31 * 2 + 28 + 30 + smallDate.m_nDay + bSm_Run; break;
    case 6:iSmdays = 31 * 3 + 28 + 30 + smallDate.m_nDay + bSm_Run; break;
    case 7:iSmdays = 31 * 3 + 28 + 30 * 2 + smallDate.m_nDay + bSm_Run; break;
    case 8:iSmdays = 31 * 4 + 28 + 30 * 2 + smallDate.m_nDay + bSm_Run; break;
    case 9:iSmdays = 31 * 5 + 28 + 30 * 2 + smallDate.m_nDay + bSm_Run; break;
    case 10:iSmdays = 31 * 5 + 28 + 30 * 3 + smallDate.m_nDay + bSm_Run; break;
    case 11:iSmdays = 31 * 6 + 28 + 30 * 3 + smallDate.m_nDay + bSm_Run; break;
    case 12:iSmdays = 31 * 6 + 28 + 30 * 4 + smallDate.m_nDay + bSm_Run; break;
    default:
        break;
    }

    //bSm_Run = ((smallDate.m_nYear % 4 == 0 && smallDate.m_nYear % 100 != 0) || smallDate.m_nYear % 400 == 0);//小年是闰年
    bBi_Run = ((bigDate.m_nYear % 4 == 0 && bigDate.m_nYear % 100 != 0) || bigDate.m_nYear % 400 == 0);
        switch (bigDate.m_nMonth){
        case 1:iBidays = bigDate.m_nDay; break;
        case 2:iBidays = 31 + bigDate.m_nDay; break;
        case 3:iBidays = 31 + 28 + bigDate.m_nDay + bBi_Run; break;
        case 4:iBidays = 31 * 2 + 28 + bigDate.m_nDay + bBi_Run; break;
        case 5:iBidays = 31 * 2 + 28 + 30 + bigDate.m_nDay + bBi_Run; break;
        case 6:iBidays = 31 * 3 + 28 + 30 + bigDate.m_nDay + bBi_Run; break;
        case 7:iBidays = 31 * 3 + 28 + 30 * 2 + bigDate.m_nDay + bBi_Run; break;
        case 8:iBidays = 31 * 4 + 28 + 30 * 2 + bigDate.m_nDay + bBi_Run; break;
        case 9:iBidays = 31 * 5 + 28 + 30 * 2 + bigDate.m_nDay + bBi_Run; break;
        case 10:iBidays = 31 * 5 + 28 + 30 * 3 + bigDate.m_nDay + bBi_Run; break;
        case 11:iBidays = 31 * 6 + 28 + 30 * 3 + bigDate.m_nDay + bBi_Run; break;
        case 12:iBidays = 31 * 6 + 28 + 30 * 4 + bigDate.m_nDay + bBi_Run; break;
        default:
            break;
        }
        if (smallDate.m_nYear == bigDate.m_nYear){
            days = iBidays - iSmdays;
        }
        else{
            days = days + 365 + bSm_Run - iSmdays;
            days = days + iBidays;
        }
        return days;
}

// DATE.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include "CDate.h"

int _tmain(int argc, TCHAR* argv[])
{
    CDate SmallDate(2007,3,12);
    CDate bigDate(2013,6,16);
    cout << DaysMinus(SmallDate, bigDate) << endl;
    system ("pause");
    return 0;
}

这是原来的一个课堂作业，可能不能满足你的要求，大致就是这么个算法。如果有用请点击下采纳。

判闰，数数就这些啊
给个比较简单的
算好闰年
先按每月31 天算
假设 d=100
先计算好每月第一天是一年第几天
100 / 31 =3.。。。 7,
4，5 两个月就好
大于 5月直接减去 5月份的第一天（+1）
不大于5 减去4月的
也可以二分查找

ccf 日期问题 求答案求答案

ccf 日期问题求答案求答案