Fortran练习经典题目.doc

a, 输入一个年、月、日并计算它是本年度的第几天。b.输入任意一年份，给出该年出生人的属相，如1945年出生的人的属相为“鸡”。c.显示输出20002099年的任何一年的某一月的月历，所要显示的年月有键盘输入。如2002年5月的月历形式如下； 5月 2002年日 一 二 三 四 五 六 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 d.显示输出20002099年的任何一年的年历，并写入一文本文件中。e.以上功能都应从菜单中选择执行。提示：a中若y为年份，d是该年的某月某日从1月1日开始经过的天数，则s=（y-1）*1.2425+d。若s除以7后所得的余数取整后为0，则为周日，为1则周一。program firstquestion implicit none integer:q,i,lday integer:year,month,day,days,mday,ds character(len=2),dimension(1:37):tem! 这个程序可以同时在屏幕和文本中输出计算结果。为了计算不同的问题，引入变量q，当q1时，计算某年月日是此年的第几天;! 当q2时，计算某年的属相;当q3时，计算月历;当q4时，计算年历;当q0时，退出此程序。此外，此程序还设置了对输入! 年、月、日的判断，判断其是否合法。 print *,when q=1,this program can resolve the question that which day some day is in some year! this is the question one. print *,when q=2,this program can obtain the shengxiao of some year! this is the question two. print *,when q=3,this program can obtain the calendar of some month of some year(2000-2099)! this is the question three. print *,when q=4,this program can obtain the Almanac of some year(2000-2099)! this is the question four. print *,If you want to exit from this program,please input q=0. do print *,input the number of q: read *,q if(q=0) exit if (q=1) then2000print *,input the year, the month and the day:read *,year,month,day if (month12.or.month12.or.month2099.or.year2099.or.year12.or.month12.or.month2099.or.year2099.or.year=12.or.m=-12) then n=mod(m,12) else n=m end ifopen(26,file=shengxiao.dat,status=unknown) if (n=0) then print *,The shengxiao of this year is:鸡 write(26,*)The shengxiao of this year is:鸡 else if(n=1.or.n=-11) then print *,The shengxiao of this year is:狗 write(26,*)The shengxiao of this year is:狗 else if(n=2.or.n=-10) then print *,The shengxiao of this year is:猪 write(26,*)The shengxiao of this year is:猪 else if(n=3.or.n=-9) then print *,The shengxiao of this year is:鼠 write(26,*)The shengxiao of this year is:鼠 else if(n=4.or.n=-8) then print *,The shengxiao of this year is:牛 write(26,*)The shengxiao of this year is:牛 else if(n=5.or.n=-7) then print *,The shengxiao of this year is:虎 write(26,*)The shengxiao of this year is:虎 else if(n=6.or.n=-6) then print *,The shengxiao of this year is:兔 write(26,*)The shengxiao of this year is:兔 else if(n=7.or.n=-5) then print *,The shengxiao of this year is:尨 write(26,*)The shengxiao of this year is:尨 else if(n=8.or.n=-4) then print *,The shengxiao of this year is:蛇 write(26,*)The shengxiao of this year is:蛇 else if(n=9.or.n=-3) then print *,The shengxiao of this year is:马 write(26,*)The shengxiao of this year is:马 else if(n=10.or.n=-2) then print *,The shengxiao of this year is:羊 write(26,*)The shengxiao of this year is:羊 else if(n=11.or.n=-1) then print *,The shengxiao of this year is:猴 write(26,*)The shengxiao of this year is:猴 end if close(26)end subroutine shengxiaosubroutine nyli(year,month,mday,ds,tem) implicit none integer:year,month,days,mday integer:i,m,ds,dts,q real:s character(len=2),dimension(1:37):tem call numday(year,month,1,days) call monday(year,month,mday) s=(year-1)*1.2425+days ds=int(s-int(s/7)*7) do i=1,37 tem(i)= end do tem(ds+1)=1;tem(ds+11)=11;tem(ds+12)=12; tem(ds+2)=2;tem(ds+13)=13;tem(ds+14)=14; tem(ds+3)=3;tem(ds+15)=15;tem(ds+16)=16; tem(ds+4)=4;tem(ds+17)=17;tem(ds+18)=18; tem(ds+5)=5;tem(ds+19)=19;tem(ds+20)=20; tem(ds+6)=6;tem(ds+21)=21;tem(ds+22)=22; tem(ds+7)=7;tem(ds+23)=23;tem(ds+24)=24; tem(ds+8)=8;tem(ds+25)=25;tem(ds+26)=26; tem(ds+9)=9;tem(ds+27)=27;tem(ds+28)=28; tem(ds+10)=10;tem(ds+29)=29;tem(ds+30)=30;tem(ds+31)=31; print (26x,i2,月,3x,i4,年),month,year print (18x,7a4),日,一,二,三,四,五,六 print (18x,7a4),(tem(i),i=1,mday+ds) print *end subroutine nylisubroutine numday(year,month,day,days) implicit none integer:year,month,day,days,n logical:leapcall leapyear(year,leap)n=int(month/2) if (month2) then if (leap) then days=days-1 else days=days-2 end if end ifend subroutine numdaysubroutine monday(year,month,mday) implicit none integer:year,month,mday logical:leap call leapyear(year,leap) select case (month) case (4,6,9,11) mday=30 case (1,3,5,7,8,10,12) mday=31 case (2) for_feb:select case (leap) case (.true.) mday=29 case (.false.) mday=28 end select for_feb case default end select end subroutine mondaysubroutine leapyear(year,leap) implicit noneinteger:yearlogical:leapif (mod(year,4)/=0) then leap=.false. else if (mod(year,100)/=0) then leap=.true. else if (mod(year,400)/=0) then leap=.true. else leap=.false.end ifend subroutine leapyearsubroutine leapday(year,month,day,lday) implicit none integer:year,month,day,lday logical:leap call leapyear(year,leap) select case (month) case (4,6,9,11) if (day30.or.day31.or.day29.or.day28.or.day1) then lday=0 else lday=1 end if end select for_feb case default end select end subroutine leapday素数是只能被1与其本身整除的数，下面给出一种寻找素数的方法：a生存一个数组，将其所有的元素初始化为1b 从数组下标为2的元素开始，每次寻找数值为1 的数组元素。当找到时，对其后的元素进行循环，凡是下标值为该元素下标值的整数倍的元素，其值为0。C当这个过程结束时，值为1的元素下标即为所求的素数。program sushu implicit none ! This program is for finding out all the prime ! number which is less than some natural number,! the result display on the screem and the text. integer,dimension(:),allocatable:shuzu integer:i,n print *,input the largest natural number: read *,n allocate(shuzu(1:n) do i=1,n shuzu(i)=1 end do call zhaosushu(shuzu,n) open(28,file=sushu.dat,status=unknown) do