给个 简单的思路

[lijianbin9]

要求 算总天数，在2008.1.1-2008.9.30之间中 输入任意日期，得出1.1号到那一天的 天数，，谢谢，，，

[oLINo]

你就算这段时间的，那简单啊，每个月的天数放到数组里，给出一个日期，得到哪月，和这个月的天数，
到数组里计算出这个月之前的天数之和，加起来就OK了

[kkndciapp]

分区段月里的天数相加

[Haimiao]

很多库里面提供了这些解法。
比如MFC，和BOOST里面的datetime。
你可以参考一下MSDN或者相关文档，再读一下源码。

[chlaws]

#include<stdio.h>
#include<string.h>
#include<conio.h>
struct  months{ 
        char   name[10];       /*月份名  */
        char   abbrev[4];      /*月份缩写 */
        int    days;           /*月份天数*/
        char   monumb[3];      /*月份号*/
}mon[12]={{"January","Jan",31,"1"},{"February","Feb",28 ,"2"},{"March","Mar",31,"3"},
{"April", "Apr", 30, "4"},{"May", "may", 31, "5"},{"June", "Jun", 30, "6"},
{"July","Jul",31,"7"},{"August", "Aug", 31, "8"},{"September", "Sep", 30, "9"},
{"October", "Oct", 31, "10"},{"November","Nov", 30 ,"11"},{"December","Dec",31,"12"}};
int func(int day,char m[],int year)
{
    int s=0,i,j,n;
    if( ((year%4 == 0) && (year%100 != 0)) || (year%400 == 0) )
	{
		mon[1].days=29;
	}
    for(i=0; i<12; i++)
    {
        if(strcmpi(mon[i].name,m)==0 || 
           strcmpi(mon[i].abbrev,m)==0 || 
           strcmpi(mon[i].monumb,m)==0)
        {
            n=i;
			//printf("i=%d\n",i);
            break;
        }
    }
    for(j=0; j<n; j++)
    {
        s+=mon[j].days;
    }
    s+=day;	
	return s;
}
void main()
{
    int sum,day,year;
    char m[20];
    printf("用户键入日号:\n");
    scanf("%d",&day);
    printf("用户键入月:月份可以是月份号、月份名或月份缩写:\n");
    scanf("%s",m);
    printf("用户键入年:\n");
    scanf("%d",&year);
    sum=func(day,m,year);
    printf("当前输入的日号是%d年的第%d天",year,sum);
	getch();
}

/*
result:
用户键入日号:
3
用户键入月:月份可以是月份号、月份名或月份缩写:
10
用户键入年:
2008
当前输入的日号是2008年的第277天
*/

[jia_xiaoxin]

如果选择的是8.8那么将1-8月的每月的天数相加然后再加上8,就得出了总天数

int main()
{
MonthDay[] = {31,29,31,30,31,30,31,31,30,31,30,31};
int month,day;
cout << "请输入月份:" << endl;
cin >> month;
cout << "请输入日期:" << endl;
cin >> day;

int allday = 0;
for(int i = 0; i < month - 1; i++)
{
allday += MonthDay[i];
}
allday += day;

cout << "总天数为:" << allday;
}

[mabo321]

求 某年某月 天数的 函数。。。

int YearMon(int year,int mon)
{
	int mday=0;
	switch(mon)
	{
	case 1: case 3: case 5: case 7: case 8: case 10: case 12:
		mday=31;
		break;
	case 2:
		if(year%4 ==0 && year%100 != 0 || year%400==0)
		{
			mday=29;
		}
		else
		{
			mday=28;
		}
		break;
	case 4: case 6: case 9: case 11:
		mday=30;
		break;
	}
	return mday;
}

[study_live]

#include "stdafx.h"
#include <iostream>
#include <time.h>

using namespace std;

int main(int argc, char* argv[]) 
{ 
	int yy, mm, dd;
	cout<<"请输入年份:";
	cin>>yy;
	cout<<"请输入月份:";
	cin>>mm;
	cout<<"请输入日期:";
	cin>>dd;

	yy-=1900;
	mm--;

	tm   t1 = {  0,   0,   0,   1,   0,   108};
	tm   t2 ;  
	t2.tm_year = yy;
	t2.tm_mon= mm;
	t2.tm_mday = dd;
	t2.tm_hour = 0;
	t2.tm_min = 0;
	t2.tm_sec = 0;

	std::cout   <<   ((mktime(&t2)   -   mktime(&t1))/(24*60*60))+1   <<   std::endl;   

	return 0; 
} 

[lijianbin9]

不错，，考虑个合适的方案，，谢谢
楼上的 辛苦了

[Walker-cheng]

[Quote=引用 1 楼 aozhi 的回复:]
参看time.h
[/Quote]
up

[aozhi]

参看time.h

[guizi110]

针对2008.1.1--2008.9.30之间的天数计算:

#include <iostream.h> 

int main() 
{ 
    int month,day; 
	int allday = 0; 
	int daymax;
	
    cout << "请输入月份:" << endl; 
    cin >> month; 
	
	while (1)
	{
		if (month>9 || month<1) 
		{
			cout << "输入月份不对，请重新输入:" << endl; 
			cin >> month; 
		}
		else
			break;
	}
	
	switch(month) 
	{
	case 1:
		daymax = 31;
		break;
	case 2:
		daymax = 29;
		allday = 31;
		break;
	case 3:
		daymax = 31;
		allday = 60;
		break;
	case 4:
		daymax = 30;
		allday = 91;
		break;
	case 5:
		daymax = 31;
		allday = 121;
		break;
	case 6:
		daymax = 30;
		allday = 152;
		break;
	case 7:
		daymax = 31;
		allday = 182;
		break;
	case 8:
		daymax = 31;
		allday = 213;
		break;
	default:
		daymax = 30;
		allday = 244;
	} 
	cout << "请输入日期:" << endl; 
    cin >> day; 
	
    while (1)
	{
		if (day>daymax || day<1) 
		{
			cout << "输入日期不对，请重新输入:" << endl; 
			cin >> day; 
		}
		else
			break;
	}
    allday += day-1; //注意此处应该减 1
	
    cout << "总天数为:" << allday; 
}

[lijianbin9]

不用time。h能有吗，，
现在还没有学到那里，，享用switch case做

[htqx]

算好了查表

[budweiser]

[Quote=引用 5 楼 chlaws 的回复:]
C/C++ code#include<stdio.h>#include<string.h>#include<conio.h>structmonths{charname[10];/*月份名*/charabbrev[4];/*月份缩写*/intdays;/*月份天数*/charmonumb[3];/*月份号*/}mon[12]={{"January","Jan",31,"1"},{"February","Feb",28,"2"},{"March","Mar",31,"3"},
{"April","Apr",30,"4"},{"May","may",31,"5"},{"June","Jun",30,"6"},
{"July","Jul",31,"7"},{"August","Aug",31,"8"},{"September","Sep",30,"9"},
{"Oc…
[/Quote]
这肯定不行嘛，平年闰年都没有考虑
可以用<time.h>

[wb186]

学习了

[zhuwanglove]

国庆回来，第一次回贴。

[xuxian02092213]

各位也忒猛了，各式各样的代码都贴出来了，真是不简单哦。。。。

这小LZ可有的看了。 参照time（）还是很好做的，time（）算的时间是从1970年1月1号开始算起的，讲两个时间一见就是中间

时间了，O(∩_∩)O~ 有所获，就必须有所悟啊。。。。

[lixiao0808120691]

UP

[aaajj]

数