C语言实验——数日子

#include <iostream.h>

struct

{ int year

int month

int day

}date

int main()

{int days

cout<<"input year,month,day:"局液

char ch='\\'

cin>>date.year>>ch>>date.month>>ch>>date.day

switch(date.month)

{ case 1: days=date.day break

case 2: days=date.day+31 break

case 3: days=date.day+59 break

case 4: days=date.day+90 break

case 5: days=date.day+120 break

case 6: days=date.day+151 break

case 7: days=date.day+181 break

case 8: days=date.day+212 break

case 9: days=date.day+243 break

case 10: days=date.day+273break

case 11: days=date.day+304break

case 12: days=date.day+334break

}

if ((date.year %4== 0 &&date.year % 100 != 0

||date.year % 400 == 0) &&date.month >=3)

days+=1

cout<<date.month<<"/"<<date.day<桐简物<" is the "咐慧<<days

<<"th day in "<<date.year<<"."<<endl

return 0

}

1970 年以后的时间，可以用 time.h 里的函数计算。时间精度为秒。按题目要求，输出时间单位用天。程序如下：

#include <stdio.h>

#include <time.h>

time_t YMD_hhmmss_2_s70(int Y, int M, int D, int hh, int mm, int ss){

struct tm *target_tm

time_t tt

time (&tt)

target_tm=localtime(&tt)

target_tm->tm_year = Y - 1900

target_tm->tm_mon= M - 1

target_tm->tm_mday = D

target_tm->tm_hour = hh // hour

target_tm->tm_min = mm

target_tm->tm_sec = ss

tt = mktime(target_tm) // from tm to time_t (s)

return tt

}

int main()

{

int y1,m1,d1,y2,m2,d2

time_t t1,t2

int dt

printf("input y1 m1 d1: ")

scanf("知并%d %d %d",&y1,&m1,&d1)

printf("\ninput y2 m2 d2: ")

scanf("%d %d %d",&y2,&m2,&d2)

t1 = YMD_hhmmss_2_s70(y1,m1,d1,0,0,0)

t2 = YMD_hhmmss_2_s70(y2,m2,d2,0,0,0)

dt = (t2-t1)/(24*3600)

printf("\ndt=%d\n",dt)

return 0

}

这里未包含日期的合法性判断。磨陵

1970 年以前 要另写程序。某年的日子是当年的第几天可用下面函数得出：

int YMD_2_JD(int Y, int M, int D){

const short MonthDay[]={0,31,28,31,30,31,30,31,31,30,31,30,31}

int JD,i

JD=D

for (i=0i<Mi++) JD+=MonthDay[i]

if (((Y%4==0)&&(Y%100!=0)||(Y%400==0)) &&(M>2)) JD++

return JD

}

整年的天数，涉及闰年的判断：搭游迹

某年是否闰年，用 (Y%4==0)&&(Y%100!=0)||(Y%400==0) 判断。闰年366天，平年365天。 有了这些，写程序不难。

未考虑公元前的年月日计算。

#include <stdio.h>

int main()

{

int year,month,day//年态滚月日

int judge,i

int sum=0//标记天数

int date[2][12]={

{31,28,31,30,31,30,31,31,30,31,30,31},

{31,29,31,30,31,30,31,31,30,31,30,31}}//储存闰年和非闰年每一月的天数

printf("输入年月日:")

scanf("%d %d %d",&year,&month,&day)

judge=(year%4==0)||(year%400==0 &&year%100!=0)

//判断年份是不是闰年帆喊余 是闰年judge就渗雀是1否则就是0

for(i=0i<month-1i++)//加month之前的所有天数

sum+=date[judge][i]

sum+=day//加当天月份的天数day

printf("它是%d年中的第%d天\n",year,sum)

return 0

}

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