给出年月日，计算该日是该年的第几天

看看是不是闰年就行了  然后把1 3 5 7 8 10 12这么月份设计为31天 由于二月不同

打两个表就行了

前几天刚写了个 你参考了：
#include <stdio.h>


int leap(int);
int exclude(int, int, int);
int calculate_days(int, int, int);


int  main( )
{
    int year, month, day, sumdays;
   
    printf("Please input year, month, day:");
    scanf("%d%d%d", &year, &month, &day);
    if (exclude(year, month, day))
    {
        sumdays = calculate_days(year, month, day);
        printf("Today is the %dth day in the year.", sumdays);
    }
    else
    {
        printf("Data error! Please enter again:");
            scanf("%d%d%d", &year, &month, &day);
    }
   
    return 0;
}

int leap(int year)
{
    if ((0 == year % 4 && 0 != year % 100) || 0 == year % 400 )
        return 1;
    else
        return 0;
}

int exclude(int year, int month, int day)
{
    int flag ;
    if (year <= 0 || month > 12 || month < 1 || day > 31 || day < 1)
        flag = 0;
    else
    {
        switch (month)
        {
        case 4:
        case 6:
        case 9:
        case 11:  if (day > 30) flag = 0; break;
        case 2:   if (leap(year) && day > 29)             { flag = 0;  break; }
                  else if (!leap(year) && day > 28)       { flag = 0;  break; }
        default:  flag = 1; break;
        }
    }
    return (flag);
}

int calculate_days(int year, int month, int day)
{
    int a[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int i, sum = 0;
   
    for (i = 1; i < month; i++)
        sum += a[i];
        sum += day;
    if (leap(year) && month >= 3)
        sum = sum + 1;
    return (sum);
}

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define SIZE 20
int count_days(struct input n);
int judge(char month[],char str1[],char str2[],char str3[]);
struct input
{
    int year;
    char month[SIZE];
    int day;
};
void main()
{
    struct input n;
    int num;
    printf("Input year,please!\n");
    scanf("%d",&n.year);
    printf("Input month,please!\n");
    fflush(stdin);
    gets(n.month);
    printf("Input day,please!\n");
    scanf("%d",&n.day);
    num=count_days(n);
    printf("%d\n",num);
    system("pause");
}
int count_days(struct input n)
{
    int all_year1[12]={31,28,31,30,31,30,31,31,30,31,30,31};
    int all_year2[12]={31,29,31,30,31,30,31,31,30,31,30,31};
    int *p,i;
    int number=0,mon,numbers;
    if(n.year%4==0&&n.year%100!=0||n.year%400==0)
        p=all_year2;
    else
        p=all_year1;
    if(judge(n.month,"1","一月","Jan"))
        mon=1;
    else if(judge(n.month,"2","二月","Feb"))
        mon=2;
    else if(judge(n.month,"3","三月","Mar"))
        mon=3;
    else if(judge(n.month,"4","四月","Apr"))
        mon=4;
    if(judge(n.month,"5","五月","May"))
        mon=5;
    if(judge(n.month,"6","六月","Jun"))
        mon=6;
    if(judge(n.month,"7","七月","Jul"))
        mon=7;
    if(judge(n.month,"8","八月","Aug"))
        mon=8;
    if(judge(n.month,"9","九月","Sep"))
        mon=9;
    if(judge(n.month,"10","十月","Oct"))
        mon=10;
    if(judge(n.month,"11","十一月","Nov"))
        mon=11;
    if(judge(n.month,"12","十二月","Dec"))
        mon=12;
    for(i=0;i<mon-1;i++)
        number+=p[i];
    numbers=number+n.day;
    return numbers;
}
int judge(char month[],char str1[],char str2[],char str3[])
{
    if(strcmp(month,str1)==0||strcmp(month,str2)==0||strcmp(month,str3)==0)
        return 1;
    else
        return 0;
}

收！

先判断是否是闰年，   然后直接加啊！！！

咦，我刚搞了一个很笨的！我是新手啊，没学几天C，但也不怕被高手们笑话了，发上来看看，希望高手指导一下啊！

/*
题目：输入某年某月某日，判断这一天是这一年的第几天？
程序分析：以3月5日为例，应该先把前两个月的加起来，然后再加上5天即本年的第几天，特殊情况，闰年且输入月份大于3时需考虑多加一天。
*/
#include<stdio.h>
#include<windows.h>
int main(void)
{
    

    int a,i=300,d,m,y,j,days;
    do{

    printf("请输入年份:");
    scanf("%d",&y);
    do
    {
    printf("请输入月份:");
    scanf("%d",&m);
    if(m>12||m==0)
    {
        printf("不好意思!您的月份输错了，一年只有12个月哟，亲!\n");
        Beep(1000,1050);
    }
    }
    while(m>12||m==0);
    do
    {
    printf("请输入日期:");
    scanf("%d",&d);
    if(((y%4==0||(y%4==0&&y%100!=0))&&(m==2))&&(d>29))
    {
        printf("您输入的日期有误!\n");
        Beep(1000,1000);
        j=1;
    }
    else if((!(y%4==0||(y%4==0&&y%100!=0)))&&(m==2)&&(d>28))
    {
        printf("您输入的日期有误!\n");
        Beep(1000,1000);
        j=1;
    }
    else if(((m==1||m==3||m==5||m==7||m==8||m==10||m==12)&&(d>31)))
    {
        printf("您输入的日期有误!\n");
        Beep(1000,1000);
        j=1;
    }
    else if((m==2||m==4||m==6||m==9||m==11)&&(d>30))
    {
        printf("您输入的日期有误!\n");
        Beep(1000,1000);
        j=1;
    }
    else
   

        j=0;
    }
    while(1==j);
    {
       

        switch(m)
    {
    case 1:
             a=0;
        break;
    case 2:
             a=31;
        break;
    case 3:
             a=59;
        break;
    case 4:
             a=90;
        break;
    case 5:
             a=120;
        break;
    case 6:
             a=151;
        break;
    case 7:
             a=181;
    case 8:
             a=212;
        break;
    case 9:
             a=243;
        break;
    case 10:
             a=273;
        break;
    case 11:
             a=304;
        break;
    case 12:
             a=334;
        break;
    default:
        break;       

    }
    if(y%4==0||(y%4==0&&y%100!=0))
    {
        if(m>2)
        {
        days=a+d+1;
        printf("\n %d 年!本年是闰年，共366天!您输入的日期是本年的第 %d 天!\n",y,days);
        }
        else
        {
            days=a+d;
            printf("\n %d 年!本年是闰年，共366天!您输入的日期是本年的第 %d 天!\n",y,days);
        }
    }
    else
    {
        days=a+d;
        printf("\n %d 年!本年是平年，共365天!您输入的日期是本年的第 %d 天!\n",y,days);
    }
    }

    printf("是否继续使用呢?继续请按1(否则按其它数字):");
    scanf("%d",&j);
    }
    while(1==j);
}

噢，我还加了些没用的报警声音，本来就是练习的嘛！


[ 本帖最后由 善水盈渊 于 2011-12-25 12:17 编辑 ]