FORTRAN經典入門程序20例

1、精品對于FORTR顏初學者。這些例子可作為小練習。1 .例題：計算工資問題。每小時工資為RATE,如果工作超過40小時，加班呢部分工資是正常時間工資的1.5倍。C Payroll with overtime program payroll real rate, hours, pay read (*,*) rate, hours if (hours>40) then pay=40*rate+(hours-40)*1.5*rate elsepay=hours*rateEND IFprint *, "rate=" , rateprint *, "hours=&qu

2、ot; , hours print *, "pay=" ,payend2 .學生成績問題。大于80為A級。大于60小于80為B級。小于60為C級。IF的嵌套。注意空格可以看清楚 else if ,end if,print 的內容.PROGRAM GRADESTUDENTREAD *,GRADEIF (GRADE .GE. 80) THENPRINT *,GRADE, "GRADE=>A"ELSEIF (GRADE.LT.60) THENPRINT*,GRADE, "GRADE=>C"ELSEPRINT*,GRADE, &q

3、uot;GRADE=>B"END IFEND IFEND3 .三個數(shù)按從小到大排序。PROGRAM MAXMINREAL A,B,C,TREAD *,A,B,CIF (A.GT.B) THENT=AA=BB=TELSEEND IFIF (B.GT.C) THENT=BB=CC=TELSEEND IFIF (A.GT.B) THENT=AA=BB=TEND IFPRINT *,A,B,CEND4 .運用EISE IF語句。重做例子2PROGRAM ex2READ (*,*) GRADEIF (GRADE .GE. 80.0) THENPRINT *, GRADE, "=

4、>A"ELSE IF(GRADE .GE. 70.0) THENPRINT *, GRADE, "=>B"ELSE IF(GARDE .GE. 60.0) THENPRINT *, GRADE, "=>C"ELSEPRINT *, GARDE, "=>D"END IFEND3x 6,x 05 .計算y 2x 2x 8,x 0PROGRAM EQUATIONREAD (*,*) XIF (X .GE. 0.0) Y=3*X+6IF (X .LT. 0.0) Y=-X*2+2*X-8PRINT *, &q

5、uot;X=" ,X, "Y=" ,YEND6 .CONTINUE語句。為了使循環(huán)的起止范圍清晰，使用終端語句例：求5!program ex1FACT=1do 10, N=1,5,1#DO 語句標號，循環(huán)不變量=初值，終值，步長FACT=FACT*N10 continueprint *,FACTend求 1 ! +2 ! +3 ! + +N -!READ *,NFACT=1sum=0.0do 10, I=1,N,1FACT=FACT*ISUM=SUM+FACT10 continueprint *,SUMend7.求 EAX=1+X+XA2/2!+X2/3!+XAN

6、/N!READ *,X,Nsum=1TERM=1do 10, I=1,N,1TERM=TERM*X/ISUM=SUM+TERM10 continueprint *,"EXP(" ,X,")" ,SUMend8.打印九九乘法表do 10,I=1,9,1do 20,J=1,IPRINT*,I,"*" ,J,"=" ,I*J20 CONTINUE10 CONTINUEend9.百錢買百雞公雞5元每只。母雞3元每只，小雞1元三只。求100元買一百只雞。INTEGER X,Y,ZDO 10,X=0,100DO 20,Y=0,

7、100DO 30,Z=0,100IF(X+Y+Z .EQ. 100) THENIF(15*X+9*Y+Z .EQ. 300) PRINT *,X,Y,ZEND IF30CONTINUE20 CONTINUE10 CONTINUEEND10.DO循環(huán)用來處理循環(huán)次數(shù)確定的情況比較方便。在處理無法事先知道循環(huán)次數(shù)時，使用當型循環(huán)。(1)DO WHILEDO 標號，WHILE(邏輯表達式)循環(huán)體CONTINUE 標號例 統(tǒng)計若干同學的得分并計算平均值。INTEGER SCORE,COUNTREAD*,SCORESUM=0COUNT=0DO 10, WHILE (SCORE .GT. 0.0)SUM=

8、SUM+SCORECOUNT=COUNT+1READ *,SCORE10 CONTINUEAVER=SUM/COUNTPRINT *,AVEREND11 .輾轉相除法求最大公約數(shù)。如26和6的最大公約數(shù)。先用27除以6 ,余數(shù)是3 ,再用6除以3 , 余數(shù)為0,則3就是最大公名數(shù)。命令 MOD()求余數(shù)。INTEGER M,N,R,TREAD*,M,NIF (M .LT. N) THENT=MM=NN=TEND IFR= MOD (M,N)DO 10, WHILE (R .NE. 0)M=NN=RR= MOD (M,N)10 CONTINUEPRINT *,"zui da gong

9、yue shu" ,NEND12.判斷一個數(shù)是否是素數(shù)。SQRT ()開根號READ *,NMAX= sqrt (real (N)I=2DO 10, WHILE (I .LE. MAX .AND. MOD (N,I) .NE. 0)I=I+110 CONTINUEIF(I .GT. MAX) THENPRINT*,N, "IS A PRIME NUMBER"ELSEPRINT *,N, "IS NOT A PRIME NUMVER"END IFEND13.求100和200之間能被3或7整除的自然數(shù)。I=10010 IF(I .LE. 200)

10、THENIF (MOD (I,3) .EQ. 0 .OR. MOD (I,7) .EQ. 0) THENPRINT *,I, "ABLE"ELSEPRINT *,I, "UNABLE”END IFI=I+1GOTO 10END IFEND14.直到型循環(huán)，UNTIL語句實現(xiàn)直到型循環(huán)。格式 “ DO S1,UNTIL(條件)循環(huán)體標號 S1 CONTINUE例14用直到型循環(huán)求5!FORTRAN不支持UNTIL語句。INTEGER I,PRODUCTI=1PRODUCT=1DO 10, UNTIL (I .GT. 5)PRODUCT=PRODUCT*II=I+11

11、0 CONTINUEPRINT *,"5!=" 'PRODUCTENDFORTRAN不支持UNTIL語句。讓我們改成 DO WHILE格式INTEGER I,PRODUCTI=1PRODUCT=1DO 10, WHILE (I .LT. 5)PRODUCT=PRODUCT*II=I+110 CONTINUEPRINT *,"5!=" 'PRODUCTEND-可編輯-15.心花矽 的雷木仙花數(shù)是瘠一土三位數(shù).再各位我宇INTEGER R,I,J,S,T,SUMDO 10, I=100,999,1T=ISUM=0DO 20 ,J=1,3,1R

12、= MOD (T,10)SUM=SUM+R*3T=T/1020 CONTINUEIF (SUM .EQ. I) PRINT *,I,"is a daffodil NUMBER"10 CONTINUEEND2_15.求一元二次方程的根Ax Bx C 0變的類型嫉肺的喇INTttqntision, KJXO-. tLfitleiiREALCXIPI5rgf*海qm KINIXLggbleALOGICALLOGrCALieipiKEian. KilNDrvHnabie)DOUBLE PRECISION A, B, CCOMPLEX X1,X2READ *,A,B,CIF(B*2-

13、4*A*C .GE.0) THENX1= CMPLX (-B+ SQRT(B*2-4*A*C)/(2*A),0.0)X2= CMPLX (-B- SQRT(B*2-4*A*C)/(2*A),0.0)ELSEX1= CMPLX (-B/(2*A), SQRT(-(B*2-4*A*C)/(2*A)X2= CMPLX (-B/(2*A),- SQRT(-(B*2-4*A*C)/(2*A)END IFPRINT *,X1,X2END16.打印此圖像。CHARACTER *9 LINEDO 10, I=-4,4,1K=1DO 40 J=1, ABS(I),1LINE(K:K尸""K=

14、K+140 CONTINUEDO 20, J=1, 2*(4- ABS(I)+1,1LINE(K:K尸"*"K=K+120 CONTINUEPRINT*,LINEDO 30, J=1,9,1LINE(J:J)=""30CONTINUE10 CONTINUEEND射垠盤一牌的修怦可 同個戳蜴便需一用只弄同 一個甲的星，伸野用中所向窗的量吊際力玷娟元*.而程序中通過設綱語向克前敲/ 例知：INl EHlIt M L m榭十都在內有中占痼一中段立刻存I*病 例4ixn <iE k al. m* z. a5.十 這六f臺即w代賽的尢個存V中無在.匕存中的位

15、道是誨跣物史，sTfflXn17.輸出數(shù)組中的數(shù)值。INTEGER A(1:10)DO 10,I=1,10,1READ *,A(I)10 CONTINUEPRINT *,AEND18.利用DO循環(huán)對數(shù)組進行輸入和輸出。從鍵盤上輸入十個數(shù)，按順序逐個放入到A(1)到A(10)中。INTEGER A(1:10)DO 10,1=1,10,1READ *,A(I)10 CONTINUEDO 20,I=1,9,2PRINT *, "A(" ,I,")” ,A(I)20 CONTINUE END19 .雙重循環(huán)將從終端讀入的數(shù)據賦值給W數(shù)組中的各元素。REAL W(1:3,1

16、:2)#表示一個三行兩列的數(shù)組，當前面數(shù)一樣時，可以寫出 W(3,2)DO 10,I=1,3,1DO 20, J=1,2,1READ *,W(I,J)20 CONTINUE10 CONTINUEPRINT *,WEND精品20。輸入10名學生的成績，統(tǒng)計各個分數(shù)段的人數(shù)。INTEGER GRADE(1:10),COUNTER(0:10)READ *,(GRADE(I),I=1,10,1)DO 10, I=1,10,1COUNTER(GRADE(I)/10)=COUNTER(GRADE(I)/10)+110 CONTINUEDO 20,1=0,10,1PRINT *,COUNTER，"=>",I*10,(I+1)*10-120 CONTINUEEND21 .順序查找。一組數(shù)據從第一個開始找，折半查找，