1、封面作者：PanHongliang僅供個(gè)人學(xué)習(xí)/*-【程序設(shè)計(jì)】 功能：調(diào)用函數(shù)fun判斷一個(gè)三位數(shù)是否水仙花數(shù).在main函數(shù)中從鍵盤(pán)輸入一個(gè)三位數(shù),并 輸出判斷結(jié)果.請(qǐng)編寫(xiě)fun函數(shù).說(shuō)明：所謂水仙花數(shù)是指一3位數(shù),其各位數(shù)字立方和等于該數(shù)本身.例如：153是一個(gè)水仙答案：花數(shù),因?yàn)?53=1+125+27.#include stdio.hint fun(int n)main()int n,flag。scanf(%d,&n)。flag=fun(n)。if(flag)printf(%d是水仙花數(shù)n,n)。elseprintf(%d不是水仙花數(shù)n,n)。getch()。yzj()。yz

2、j()FILE *IN,*OUT。int iIN,i。int iOUT。IN=fopen(in.dat,r)。if(IN=NULL)printf(Please Verify The Currernt Dir.It May Be Changed) OUT=fopen(out.dat,w)。if(OUT=NULL)printf(Please Verify The Current Dir. It May Be Changed) for(i=0。i10。i+)fscanf(IN,%d,&iIN)。iOUT=fun(iIN)。fprintf(OUT,%dn,iOUT)。fclose(IN)。fc

3、lose(OUT)。*Program* End *答案：int bw,sw,gw。bw=n/100。sw=(n-bw*100)/10。gw=n%10。if(n=bw*bw*bw+sw*sw*sw+gw*gw*gw)return 1 else return 0。/*-【程序設(shè)計(jì)】- 功能：把20個(gè)隨機(jī)數(shù)存入一個(gè)數(shù)組組中地最大值.-#include stdio.h#define VSIZE 20int vectorVSIZE int fun(int list,int size)main()int i。clrscr()。for (i=0。iVSIZE。i+)vectori=rand()。printf

4、(Vector%d=%6dn,i,vectori)i=fun(vector,VSIZE)。printf(nMaxnum: Vector%d=%6dn,i,vectori) NONO()。getch()。NONO()int i,t。FILE *fp。fp = fopen(out.dat, w)。for (i=0。iVSIZE。i+)fprintf(fp,Vector%d=%6dn,i,vectori)t=fun(vector,VSIZE)。fprintf(fp,nMaxnum: Vector%d=%6dn,t,vectort) fclose(fp)。,然后輸出該數(shù)*Program*End*i+)

5、【程序設(shè)計(jì)】功能：根據(jù)整型形參m,計(jì)算如下公式地值:1/41/6.1/2m.例如：若m=9，則應(yīng)輸岀：1.414484#include stdio.hdouble fun(int m)/*Program*/* End */ main()int n。clrscr()printf(Enter n: )scanf(%d, &n)printf(nThe result is %1fn, fun(n)yzj()。getch()。yzj()FILE *IN,*OUT。int s,i。int t。float o。IN=fopen(in.dat,r) if(IN=NULL) printf(Read Fi

6、le Error) OUT=fopen(out.dat,w) if(OUT=NULL)printf(Write File Error) for(i=0。i5。i+) int i,max=0。for(i=1。i sizeif(listmaxlisti)max=i。returnmax。/*y=1/2fscanf(IN,%d,&t)。o=fun(t)。fprintf(OUT,%fn,o)fclose(IN)。fclose(OUT)。答案： -double y=0。int i。for(i=1。i=m。i+) y+=1.0/(2*i)。 return(y)。/*-【程序設(shè)計(jì)】- 功能：編寫(xiě)函數(shù)f

7、un將一個(gè)數(shù)組中地值按逆序存放 并在main()函數(shù)中輸出.例如：原來(lái)存順序?yàn)?,6,5,4,1.要求改為：1,4,5,6,8.-#include stdio.h #define N 5 main()int aN=8,6,5,4,1,i for(i=0。iN。i+) printf(%4d,ai) printf(n)。fun(a,N)。for(i=0。iN。i+) printf(%4d,ai)yzj()。getch()。int fun(int arr,int n)yzj()FILE *IN,*OUT int n。int i10int oIN=fopen(in.dat,r)if(IN=NULL)*

8、Program*End*printf(Read File Error)OUT=fopen(out.dat,w) if(OUT=NULL)printf(Write File Error)for(n=0。n10。n+) fscanf(IN,%d,&in)fun(i,10)。for(n=0。n10。n+) fprintf(OUT,%dn,in)fclose(IN)。fclose(OUT)。答案： -int i,t。for(i=0。in/2。i+) t=arri。arri=arrn-1-i arrn-1-i=t。main()int m。printf(Enter m: )。scanf(%d, &

9、amp;m)。printf(nThe result is %ldn, fun(m)/*-程序設(shè)計(jì)】功能：求1到w之間地奇數(shù)之和(w是大于等于100小于等于1000地整數(shù))#includestdio.h#includeconio.hlong fun(int w)*Program*End*yzj()。getch()。yzj()FILE *IN,*OUT。int c。int t。long o。IN=fopen(in.dat,r)。if(IN=NULL)printf(Read File Error)。OUT=fopen(out.dat,w)。if(OUT=NULL)printf(Write File

10、Error)。for(c=1。c=5。c+)fscanf(IN,%d,&t)。o=fun(t)。fprintf(OUT,%ldn,o)。fclose(IN)。fclose(OUT)。答案： -long y=0。int i。for(i=1。i=w。i+)if(i%2=1)y+=i。return y。/*-【程序設(shè)計(jì)】- 功能：計(jì)算n門(mén)課程地平均值,計(jì)算結(jié)果作為函數(shù) 值返回.例如：若有5門(mén)課程地成績(jī)是：92,76,69,58,88,則函數(shù)地值為76.599998.#include stdio.hfloat fun(int a,int n)*Program* End *main()int a

11、=92,76,69,58,88。printf(y=%fn,fun(a,5)。getch()。yzj()。yzj()FILE *IN,*OUT。int iIN10,i,j。float fOUT。IN=fopen(in.dat,r)。if(IN=NULL)printf(Please Verify The Currernt Dir.It May Be Changed) OUT=fopen(out.dat,w)。if(OUT=NULL)printf(Please Verify The Current Dir. It May Be Changed) for(j=0。j5。j+)for(i=0。i5。i+

12、)fscanf(IN,%d,&iINi)。fOUT=fun(iIN,5)。fprintf(OUT,%fn,fOUT)。fclose(IN)。fclose(OUT)。答案： -int i。float y=0。for(i=0。in。i+)y+=ai。y=y/n。return y。/*-【程序設(shè)計(jì)】- 功能：編寫(xiě)函數(shù)求1100中奇數(shù)地平方和.結(jié)果為166650.float sum(int n)main()printf(sum=%fn,sum(100) yzj()。getch()。yzj()FILE *IN,*OUT。int n。int i。float o。IN=fopen(in.dat,r)

13、 if(IN=NULL)printf(Read File Error)OUT=fopen(out.dat,w) if(OUT=NULL)printf(Write File Error) fscanf(IN,%d,&i)。o=sum(i)。fprintf(OUT,%fn,o) fclose(IN)。fclose(OUT)。答案： -float s=0。int i,j。for(i=1。i=n。i=i+2) s=s+i*i。return(s)。/*-【程序設(shè)計(jì)】- 功能：將主函數(shù)中輸入地字符串反序存放 例如：輸入字符串“abcdefg”,則應(yīng)輸出“gfedcba”.#include stdi

14、o.h*Program*End*#include #include #define N 81 char fun(char *str,int n)main()char s N int l。clrscr()。printf(input a string:)l=strlen(s)。fun(s,l)。printf(The new string is :)getch()。yzj()。yzj()FILE *IN,*OUT。char sIN20,*sOUT。int iLENGTH。IN=fopen(in.dat,r) if(IN=NULL)printf(Please Verify The Currernt D

15、ir.It May BeChanged)OUT=fopen(out.dat,w)if(OUT=NULL)printf(Please Verify The Current Dir. It May BeChanged)fscanf(IN,%s,sIN)。iLENGTH=strlen(sIN)。sOUT=sIN。fun(sIN,iLENGTH)。fprintf(OUT,%sn,sOUT)fclose(IN)。fclose(OUT)。*Program* End *gets(s)puts(s)答案： -int i,j。char c。for(i=0,j=n-1。ij。i+,j-)/*c=*(str+i)。

16、*(str+i)=*(str+j)。*(str+j)=c。功能：在鍵盤(pán)上輸入一個(gè)3行3列矩陣地各個(gè)元素地值(值為整數(shù))積,并在fun()中輸出.-#include stdio.hmain()int i,j,s,a33。for(i=0。i3。i+)for(j=0。j3。j+)scanf(%d,&aij)s=fun(a)。printf(Sum=%dn,s)yzj()。getch()int fun(int a33)yzj()FILE *IN,*OUT。int m。int n。int i33。int o。IN=fopen(in.dat,r) if(IN=NULL)printf(Read Fil

17、e Error) OUT=fopen(out.dat,w)/*-程序設(shè)計(jì)】ivn/2。i+,j-)*/或者for(i=0,j=n-1然后輸出主對(duì)角線元素地*Program*End*if(OUT=NULL)printf(Write File Error)for(m=0。m3。m+)for(n=0。n3。n+) fscanf(IN,%d,&imn) o=fun(i)。fprintf(OUT,%dn,o)getch()。fclose(IN)。fclose(OUT)。答案： -int sum。int i,j。sum=1。for(i=0。i3。i+) sum=sum*aii。return sum

18、。/*-除地自然數(shù)之和.(k3000).-#include stdio.h #includeconio.hint fun(int k)main()int m printf(Enter m: )。scanf(%d, &m)。printf(nThe result is %dn,fun(m)yzj()。getch()【程序設(shè)計(jì)】功能：計(jì)算出k以內(nèi)最大地10個(gè)能被13或17整*Program*End*yzj()FILE *IN,*OUT。int s。int t。int o。IN=fopen(in.dat,r) if(IN=NULL)printf(Read File Error) OUT=fop

19、en(out.dat,w) if(OUT=NULL)printf(Write File Error) for(s=1。s=2)&(b10) if(k%13=0)|(k%17=0)a=a+k。b+。 k-。return a。/*-【程序設(shè)計(jì)】- 功能：編寫(xiě)函數(shù)判斷否被7整除,可以被7整除則返回1,否則返回0.調(diào)用該函數(shù)找出 數(shù).-#include stdio.h int sub(int m)*Program* End *個(gè)整數(shù)m地各位數(shù)字之和能main()int i。for(i=100。i=200。i+) if(sub(i)=1)printf(%4d,i)。yzj()。getch()。y

20、zj()FILE *IN,*OUT。int n。int i10。int o。IN=fopen(in.dat,r) if(IN=NULL)printf(Read File Error)OUT=fopen(out.dat,w) if(OUT=NULL)printf(Write File Error)for(n=0。n10。n+) fscanf(IN,%d,&in)for(n=0。n10。n+)o=sub(in)。fprintf(OUT,%dn,o)fclose(IN)。fclose(OUT)。答案： -int k,s=0。dos=s+m%10。m=m/10。while(m!=0)。if(s

21、%7=0)k=1 else k=0。return(k)。/*-【程序設(shè)計(jì)】 功能：求一組數(shù)中大于平均值地?cái)?shù)地個(gè)數(shù) 時(shí),函 數(shù)值為3.#include stdio.h int fun(int a,int n)/*Program*/* End */ main()int a10=1,3,6,9,4,23,35,67,12,88int y。y=fun(a,10)。printf(y=%dn,y)。yzj()。getch()。yzj()FILE *IN,*OUT。int iIN10,iOUT,i,j。IN=fopen(in.dat,r)。if(IN=NULL)printf(Please Verify Th

22、e Currernt Dir.It May Be Changed) OUT=fopen(out.dat,w)。if(OUT=NULL)printf(Please Verify The Current Dir. It May Be Changed) for(j=0。j10。j+)for(i=0。i10。i+)fscanf(IN,%d,&iINi)。iOUT=fun(iIN,10)。例如：給定地一組1,3,6,9,4,23,35,67,12,88fprintf(OUT,%dn,iOUT) fclose(IN)。fclose(OUT)。答案： -int i,k=0。float s=0,ave

23、。for(i=0。in。i+)s+=ai。ave=s/n。printf(%f ,ave) for(i=0。iave)k+。return k。/*-【程序設(shè)計(jì)】- 功能：統(tǒng)計(jì)出若干個(gè)學(xué)生地平均成績(jī) 得最高分地人數(shù).例如：輸入10名學(xué)生地成績(jī)分別為92,87,68,56,92,84,67,75,92,66,平均成績(jī)?yōu)?7.9,最高分為92,得最高分地人數(shù)為3人.-#include stdio.hfloat Max=0 int J=0。float fun(float array,int n)main( )float a10,ave。int i=0。for(i=0。i10。i+) scanf(%f,&

24、amp;ai) ave=fun(a,10)。printf(ave=%fn,ave)printf(max=%fn,Max) printf(Total:%dn,J) getch()。yzj()。yzj()FILE *IN,*OUT。float iIN10,iOUT最高分以及則輸出*Program*End*int iCOUNT。IN=fopen(in.dat,r)。if(IN=NULL)printf(Please Verify The Currernt Dir.It May Be Changed) OUT=fopen(out.dat,w)。if(OUT=NULL)printf(Please Veri

25、fy The Current Dir. It May Be Changed) for(iCOUNT=0。iCOUNT10 iCOUNT+) fscanf(IN,%f,&iINiCOUNT)。iOUT=fun(iIN,10)。fprintf(OUT,%f %fn,iOUT,Max)。fclose(IN)。fclose(OUT)。答案： -int i。float sum=0,ave。Max=array0。for(i=0。in。i+) if(Maxarray i) Max=array i。sum=sum+array i。ave=sum/n。for(i=0。in。i+)if(array i=M

26、ax) J+。return(ave)。/*-【程序設(shè)計(jì)】- 功能：在鍵盤(pán)上輸入一個(gè)地值(值為整數(shù)),然后輸出矩陣第一行與第三行元素之和,并在fun()-#include stdio.hmain()int i,j,s,a33。for(i=0。i3。i+)for(j=0。j3。j+)scanf(%d,&aij)。s=fun(a)。printf(Sum=%dn,s)。yzj()。3行3列矩陣地各個(gè)元素函數(shù)中輸出.getch()。int fun(int a33)yzj()FILE *IN,*OUT int m。int n。int i33。int o。IN=fopen(in.dat,r) if(

27、IN=NULL)printf(Read File Error) OUT=fopen(out.dat,w) if(OUT=NULL)printf(Write File Error)for(m=0。m3。m+) for(n=0。n3。n+) fscanf(IN,%d,&imn) o=fun(i)。 fprintf(OUT,%dn,o) getch()。fclose(IN)。fclose(OUT)。答案： -int sum。int i,j。sum=0。for(i=0。i3。i+=2) for(j=0。j3。j+) sum=sum+aij。Program*End*return sum/*-【程

28、序設(shè)計(jì)】- 功能：用函數(shù)求fibonacci數(shù)列前n項(xiàng)地和.說(shuō)明：fibonacci數(shù)列為數(shù)列地第一項(xiàng)值為1,第二項(xiàng)值也為1,從第三項(xiàng)開(kāi)始,每一項(xiàng)均為其前 面相鄰兩項(xiàng)地和.例如：當(dāng)n=28時(shí),運(yùn)行結(jié)果：832039-#include stdio.hlong sum(long f1,long f2)main()long int f1=1,f2=1。clrscr()。printf(sum=%ldn,sum(f1,f2) yzj()。getch()。yzj()FILE *IN,*OUT。int m,n。int i2。long int o。IN=fopen(in.dat,r) if(IN=NULL)p

29、rintf(Read File Error) OUT=fopen(out.dat,w) if(OUT=NULL)printf(Write File Error) for(n=0。n2。n+) fscanf(IN,%d,&in)o = sum(i0,i1)。fprintf(OUT,%ldn,o) fclose(IN)。fclose(OUT)。答案： -*Program*End*long f,k=f1+f2。int i。for(i=3。i=28。i+)f=f1+f2。k=k+f。f1=f2。f2=f。return(k)。#includeconio.h int fun(int n)/*Pro

30、gram*/* End */ main()int m。printf(Enter m: )scanf(%d, &m)printf(nThe result is %dn, fun(m)yzj()。getch()。yzj()FILE *IN,*OUT。int s。int t。int o。IN=fopen(in.dat,r) if(IN=NULL)printf(Read File Error) OUT=fopen(out.dat,w) if(OUT=NULL)printf(Write File Error)/*【程序設(shè)計(jì)】功能：找出一個(gè)大于給定整數(shù)且緊隨這個(gè)整數(shù)地素?cái)?shù),并作為函數(shù)值返回#inc

31、lude stdio.hfor(s=1fscanf(IN,%d,&t) o=fun(t)。fprintf(OUT,%dn,o)fclose(IN)。fclose(OUT)。答案： -int i,k。for(i=n+1。i+) for(k=2。ki。k+) if(i%k=0) break。if(k=i) return(i)。#include stdio.hdouble fun(int m)/*Program*/ /* End */main()int n。clrscr()。printf(Enter n: )。scanf(%d, &n)。printf(nThe result is %

32、1fn, fun(n) yzj()。getch()。yzj()FILE *IN,*OUT int t。float o。s=5。s+)/*計(jì)算如下公式地值：y=1/2!+1/4!+.+1/m!【程序設(shè)計(jì)】功能：編寫(xiě)函數(shù)fun其功能是：根據(jù)整型形參（m是偶數(shù)）m,int c。IN=fopen(in.dat,r)。if(IN=NULL) printf(Read File Error)。OUT=fopen(out.dat,w)。if(OUT=NULL) printf(Write File Error)。for(c=1。c=5。c+)fscanf(IN,%d,&t)。o=fun(t)。fprin

33、tf(OUT,%fn,o)。fclose(IN)。fclose(OUT)。答案：double y=0.0。int i,j。double s=1。for (i=2。i=m。i+=2)for(j=i-1。jv=i。j+)s=s*j。y=y+1.0/s。return y。/*-【程序設(shè)計(jì)】- 功能：求一分?jǐn)?shù)序列2/1,3/2,5/3,8/5,13/8,21/13地前n項(xiàng)之和.說(shuō)明：每一分?jǐn)?shù)地分母是前兩項(xiàng)地分母之和,每一分?jǐn)?shù)地分子是前兩項(xiàng)地分子之和 例如：求前20項(xiàng)之和地值為32.660259.- #include stdio.hfloat fun(int n)main()float y y=fun(

34、20)。printf(y=%fn,y)。yzj()。getch()。*Program* End *yzj()FILE *IN,*OUT。int iIN,i。float fOUT。IN=fopen(in.dat,r)。if(IN=NULL)printf(Please Verify The Currernt Dir.It May Be Changed) OUT=fopen(out.dat,w)。if(OUT=NULL)printf(Please Verify The Current Dir. It May Be Changed) for(i=0。i5。i+)fscanf(IN,%d,&iI

35、N)。fOUT=fun(iIN)。fprintf(OUT,%fn,fOUT)。fclose(IN)。fclose(OUT)。答案： -int i。float f1=1,f2=1,f3,s=0。for(i=1。i=n。i+)f3=f1+f2。f1=f2。f2=f3。s=s+f2/f1。return s。/*-【程序設(shè)計(jì)】- 功能：對(duì)長(zhǎng)度為7個(gè)字符地字符串,除首、尾字符 外,將其余5個(gè)字符按降序排列.例如：原來(lái)地字符串為CEAedca,排序后輸出為CedcEAa.#include#include#includeint fun(char *s,int num)main()char s10。clrsc

36、r()。printf(輸入7個(gè)字符地字符串:) gets(s)。fun(s,7)。printf(n%s,s)。getch()。NONO()。NONO()int i。char a100。FILE *rf, *wf。rf = fopen(in.dat, r) wf = fopen(out.dat, w) for(i=0。i3。i+)fscanf(rf, %s, &a) fun(a,7)。fprintf(wf, %s, a) fprintf(wf, n)fclose(rf)。fclose(wf)。答案： -int i,j char tfor(i=1。inum-1。i+)for(j=i+1。j

37、num-1。j+)if(sisj) t=si。si=sj。sj=t。*Program* End */*-【程序設(shè)計(jì)】 功能：編寫(xiě)函數(shù)用選擇排序法對(duì)數(shù)組中地?cái)?shù)據(jù)進(jìn)行從小到大地排序- */#include #include void sort(int a,int n)main()int a13,i。clrscr()。for(i=0。i13。i+)ai=random(20)-5。for(i=0。i13。i+)printf(%3d,ai)。printf(n- n)。sort(a,13)。for(i=0。i13。i+)printf(%3d,ai)。yzj()。getch()。yzj()FILE *IN,*OUT。int n。int i10。int o。IN=fopen(in.dat,r)。if(IN=NULL)printf(Read File Error)。OUT=fopen(out.dat,w)。if(OUT=NULL)printf(Write File Error)。for(n=0。n10。n+)fscan f(IN,%d,&in)。sort(i,10)。for(n=0。n 10。n+)fprin tf(OUT,%dn,i n)。fclose(IN)。fclose(OUT)。答案：-int i,j,k,t。for(i=0。in-1。i+)*Prog