心情不错，写了一个小程序玩

准确计算两个日期之间的天数，年.月.日
这个程序应该可以很简单，可是为了最完美，里面加了一些错误检查和防止键盘输入缓冲区溢出的一段代码，不知道这样实现可不可以。

文件:	date_count.rar
大小:	89KB
下载:	下载

 

#include <iostream>
using namespace std;

struct date
{
    int year;
    int month;
    int day;
}older,later;

int day[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31},
                {0,31,29,31,30,31,30,31,31,30,31,30,31}};

int account()
{
    int sum=0;
    int i=0;
    int month=0;
    int leap_o,leap_l;

    for (i=older.year ; i<later.year ;i++) //lesved middle years count

    {
        if (i%4==0 && i%100!=0 || i%400==0)
            sum+=366;
        else
            sum+=365;
    }

    leap_o=older.year%4==0 && older.year%100!=0 || older.year%400==0;
    leap_l=later.year%4==0 && later.year%100!=0 || later.year%400==0;

    if (sum!=0) //in the different years

    {
        if (later.year - older.year ==1)
        {
            for (i=1;i<older.month;i++)
                sum-=day[leap_o][i];
         sum-=older.day;

            for (i=1;i<later.month;i++)
                sum+=day[leap_l][i];
            sum+=later.day;
        }
    }
    else //in the same year

    {
        if (older.month != later.month)
        {
            for (i=older.month+1 ; i<later.month ; i++)
                sum+=day[leap_l][i];
            sum+=(day[leap_o][older.month]-older.day);
            sum+=day[leap_l][later.month];
        }
        else
            sum+=(later.day-older.day);
    }

    return sum;
}

int todate(char line[],int i)
{
    int state=0;//It shows what means about the '.'

    int pointer=0;
    int sum=0;

    while (line[pointer]!='\n')
    {
        if (line[pointer]!='.')
        {
            sum=sum*10+line[pointer]-'0';
        }
        else
        {
            switch (state)
            {
            case 0:if (i==1)
                 {
                     older.year=sum;
                     sum=0;
                 }
             else
                 {
                     later.year=sum;
                     sum=0;
                 }
                 state++;
                 break;
            case 1:    if (sum>12)
                        return -1;
                 if (i==1)
                 {
                 older.month = sum;
                 sum=0;
                 }
                 else
                 {
                     later.month = sum;
                 sum=0;
                 }
                 state++;
                 break;
            }
        }
        pointer++;
    }
    if (state == 2)
    {
        if (sum>31)
            return -1;
        if (i==1)
        {
            older.day = sum;
            sum=0;
        }
        else
        {
            later.day = sum;
            sum=0;
        }
    }
    if (state!=2)
        return -1;
    return 0;
}

int enter(char line[],int k)
{
    char c;
    int i=0;
    int dot=1;
    
    //Below I may solve the problem about keybord buffer.

    while ((c=getchar())!='\n')
    {
        if (i>10)
        {
            i=0;
        }
        line[i++]=c;
    }
    line[i]='\n';
    i=0;

    while (line[i]!='\n')
    {
        if (line[i]<='9'&&line[i]>='0'||line[i]=='.')
        {
            if (line[i]=='.')
                dot++;
            if (dot >3)
                return -1;
        }
        else
            return -1;
        i++;
    }

    i=todate(line,k);

    return i;
}

void main()
{
    char line[11];
    int sum_day=0;

    cout <<"Please enter the older date:";

    while (enter (line,1)==-1)
    {
        cout <<"Date Error!!"<<endl;
        cout <<"Please reenter:";
    }

    cout <<"Please enter the later date:";

    while (enter (line,2)==-1)
    {
        cout <<"Date Error!!"<<endl;
        cout <<"Please reenter:";
    }

    sum_day=account();
    if (sum_day<0)
        cout <<"Error!!"<<endl;
    else
        cout <<"The sum of the day is "<<sum_day<<endl;
}