大学计算机C语言程序设计(程序设计).docx

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1、一、程序设计共113题第1题题号:319/*【程序设计】功能：判断m是否为素数。*/#include stdio. h void wwjt ();int fun(int m)/*Program*/* End */ main ()int m, k=0;for(m=100;m200;m+) if(fun(m)printf(%4d”,m);k+;if(k%5=0) printf(n);)printf(k=%dn”,k); wwjt();void wwjt ()FILE *IN,*OUT; int ilN, iOUT, i;IN=fopen (*in. dat,r);if(IN=NULL)(print

2、f(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);fclose(OUT);答案：int i, k=l;if(m=l) k=0;for(i=2;im;i+)if(m%i=0) k=0;r

3、eturn k;第2题题号:351【程序设计】功能：请编一个函数 void fun(int ttMN, int ppN).tt指向一个M行N列的二维数组，求出二维数组每列中最大元素，并依次放入PP所指一维数组中。二维数组中的数已在主函数中赋予。*/#include#includettdefine M 3ttdefine N 4void wwjt();void fun(int ttMN, int ppN)/*Pr()gram*/* End */)mainOint tMN=22,45,56,30,19,33,45,38),20,22,66,40;int pN, i, j, k;printf C*T

4、he original data is:n);for(i=0;iM;i+)(for(j=0;jN;j+)printf (*%6d*, tij);printf(n);fun(t, p);printf(*nThe result is:n);for(k=0;kN;k+) printf(*%4d*, pk);printf(n);wwjt ();void wwjt ()(int i, j, array34,p4;FILE *rf,*wf ;rf = fopen(in. dat,r);wf = fopen(*out. dat*,w);for (i=0; i 3; i+)for (j=0; j 4; j+)

5、fscanf (rf,%d”,&arrayij);fun (array, p);for (j=0; j 4; j+)fprintf (wf,%7d”, pj); fprintf (wf,n);)fclose(rf);fclose(wf);答案：-int i, j;for(j=0;jN;j+)(ppj=ttoj;for(i=l;ippj)ppj=ttij;第3题题号:375/*【程序设计】功能：从键盘上输入任意实数，求出其所对应的函数值。z=e 的 x 次幕(x10)z=log(x+3)(x-3)z=sin(x)/(cos (x)+4)*/include #includevoid wjt ();

6、double y (float x)(/*Prograiii*/* End */main ()float x;scanf (*%f *,&x);printf(y=%fn, y(x);wwjt ();void wwjt ()FILE *IN,*0UT;int n;float i;double 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) z=exp(x);else if

7、(x-3) z=log(x+3);else z=sin(x)/(cos(x)+4);return (z);第4题题号:334/*【程序设计】功能：求给定正整数n以内的素数之积。(n28)*/#include #includeconio. hvoid wjt ();long fun(int n)(/*Prograni*/* End */ mainO(int m;printf(*Enter m:);scanf (*%d*,&m);printf(*nThe result is %ldn*, fun(m); wwjt ();void wwjt ()(FILE *IN,*OUT;int c ;int t

8、;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 Error*);)for(c=l;c=5;c+)(fscanf (IN,%d,&t);o=fun(t);fprintf(OUT,o);)fclose(IN);fclose(OUT);)答案：-long i, k;long s=l;for(i=2;i=n;i+)for(k=2;ki;k+)if (i%k=0) break;if(k=i)s=s*i;retu

9、rn s;第5题题号:333【程序设计】功能：计算并输出给定整数的所有因子之积（包括自身）。规定这个整数的值不大于50o*/#include #include*conio.h void wwjt ();long int fun(int n)/*Program*/main ()int m;printf(Enter m:);scanf (*%d*,&m);printf(*nThe result is %ldn* fun(m);wwjt ();)void wwjt ()FILE *IN,*OUT;int s ;int t;long int o;IN二fopen(in. dat*,r);if(IN=N

10、ULL)(printf(*Read FILE Error*);OUT二fopen (out. dat,w);if(OUT=NULL)(printfCWrite FILE Error*);)for(s=l;s=5;s+)(fscanf (IN,%d,&t);o=fun(t);fprintf (OUT,%ldn”, o);fclose(IN);fclose(OUT);答案：-long s=l, i;for(i=2;i=n;i+) if(n%i=0)s=s*i;return s;第6题题号:50/*【程序设计】功能：求出菲波那契数列的前一项与后一项之比的极限的近似值。例如：当误差为0.0001时，函

11、数值为0.618056。*/#include #include math, h”void wwjt ();float fun()/*Prograiii*/* End */)main()(printfCy=%fn*,fun();wwjt ();void wwjt ()(FILE *IN,*OUT;float fOUT;IN=fopen(*19. IN*,*r*);if(IN=NULL)(printf(*Please Verify The Currernt Dir. it May Be Changed);OUT=fopen(19. out,w);if(OUT=NULL)(printf(Please

12、 Verify The Current Dir. it May Be Changed);fOUT=fun();fprintf(OUT,%fn，fOUT);fclose(IN);fclose(OUT);float fl=l,f2=l,f3;float rl=2,r2;dor2=rl;rl=fl/f2;f3=fl+f2;fl=f2;f2=f3;while(fabs(rl-r2)le-4);return rl;第7题题号:394【程序设计】功能：产生20个30,120上的随机整数放入二维数组 a54中,求其中的最小值。-*/include *stdlib. h#include void wwjt (

13、);int amin(int a54)/*Prograni*/* End */ main()int a54, i, j;for(i=0;i5;i+)for(j=0;j4;j+)aij=random(120-30+1)+30;for(i=0;i5;i+)(for (j=0; j4; j+)printf (%5d, aij);printfCn*)；)printf(min=%dn”, amin(a);wwjt ();)void wwjt ()(FILE *IN,*OUT;int m, n;int i54;int o;IN=fopen(*in. dat,r);if(IN=NULL)(printfCRe

14、ad FILE Error*);0UT=fopen(*out. dat,w);if(OUT二二NULL)printf(*Write FILE Error*);)for(m=0;m5;m+)for(n=0;n4;n+)(fscanf (IN,*%d*,&i mn);o=amin(i);fprintf(OUT,*%dn*, o);fclose(IN);fclose(OUT);答案:int i, j, s;s=a00;for(i=0;i5;i+)for(j=0;jaij)s=aij;return(s);第8题题号:337/*【程序设计】功能：将两个两位数的正整数a、b合并形成一个整数放在 c中。合并

15、的方式是：将a数的十位和个位数依次放在c数的百位和个位上，b数的十位和个位数依次放在c数的十位和千位上。例如：当a=45, b=12o调用该函数后，c=2415o*/#include void wjt ();void fun(int af int b, long *c)/*Program*/main ()(int a, b; long c;printf(input a, b:);scanf (%d%d”,&a,&b);fun (a, b,&c);printf(*The result is:%ldn*, c);wwjt ();void wwjt ()(FILE *rf,*wf ;int i, a

16、, b ; long c ;rf = fopen(in. dat,r);wf = fopen (out. dat,w);for(i =0; i 10; i+)(fscanf (rf,%d,%d,&a,&b);fun (a, b,&c);fprintf (wf,*a=%d, b=%d, c=%ldn*, a, b, c)fclose(rf);fclose(wf);答案：*c=a/l0*100+a%10+b/l0*10+b%10*1000;第9题题号:388/*【程序设计】功能：编写函数判断一个整数能否同时被3和5整除，若能则返回值为1,否则为0。调用该函数求出15300之间能同时被3和5整除的数

17、的个数。*/#include void wwjt ();int sum(int n)(/*Prograni*/* End */)main ()(int i,s=0;for(i=15;i=300;i+)if(sum(i)=l) s=s+l;printf (*s=%dn*, s);wwjt ();)void wwjt ()(FILE *IN,*OUT;int n;int i12;int o;IN=fopen(in. dat,r);if(IN=NULL)(printfCRead FILE Error);)0UT=fopen(*out. dat,w);if(OUT=NULL)(printf(*Writ

18、e FILE Error*);for(n=0;n5;n+)fscanf(IN,%d”,&in);for(n=0;n5;n+)o = sum(in);fprintf(OUT,%dn, o);)fclose(IN);fclose(OUT);答案一if(n%3=0&n%5=0) return(l);return(0);第10题题号：16/*【程序设计】功能：编写函数fun求1!+2!+3!+n!的和，在main函数中由键盘输入n值，并输出运算结果。请编写fun 函数。例如：若n值为5,则结果为153。/#include void wwjt ();long int fun(int n)/*Progra

19、m*/* End */ main ()int n;long int result;scanf &n);result=fun(n);printf(*%ldn*, result);wwjt ();)void wwjt()(FILE *IN,*OUT;int i;int ilN;long 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 Ver

20、ify The Current Dir. It May Be Changed*);)for(i=0;i10;i+)fscanf (IN,*%d,&iIN);iOUT=fun(iIN);fprintf(OUT,%ldn”, iOUT);fclose(IN);fclose(OUT);答案：int i;float f=l, s=0;for(i=l;i=n;i+)f=f*i;s=s+f;return s;第11题题号：329【程序设计】功能：对某一正数的值保留2位小数，并对第三位进行四舍五入。I#include #includeconio. hvoid wjt ();double fun(float

21、h)(/*Prograni*/* End */ mainO(float m;printf(*Enter m:);scanf (*%f*,&m);printf(*nThe result is %8.2fn*, fun(m); wwjt ();void wwjt ()(FILE *IN,*OUT;int s ;float t;double 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(s=l;

22、s=5)return(int)(h*100+l)/100.0;elsereturn(int)(h*100)/100.0;第12题题号:410【程序设计】功能：编写函数fun将个数组中的值按逆序存放，并在 mainO函数中输出。例如：原来存顺序为8,6,5,4,1。要求改为：1,4,5,6,8。*/#include #define N 5void wwjt ();int fun(int arr, int n)/*Program*/main ()int aN=8,6,5,4,1, i;for(i=0;iN;i+)printf (%4d, ai);printf(n);fun (a, N);for(i

23、=0;iN;i+)printf (%4d, ai);wwjtO ;)void wwjt()(FILE *IN,*OUT;int n;int i10;IN=fopen(in. dat,r); if(IN=NULL)printfCRead FILE Error);0UT=fopen (out. dat,w);if(OUT=NULL)(printf(Write FILE Error*)：for(n=0;n10;n+)fscanf (IN,%d,&i n);fun(i,10);for(n=0;n10;n+)fprintf(OUT,%dn, in);fclose(IN);fclose(OUT);答案：i

24、nt i, t;for(i=0;in/2;i+) t=arri;arri=arrn-l-i; arrn-l-i=t;第13题题号：327/*【程序设计】功能：能计算从1开始到n的自然数的和，n由键盘输入,并在mainO函数中输出。*/#include void wwjt ();int fun(int n)/*Prograiii*/* End */ mainO(int m;printf(Enter m:); scanf (*%d*,&m);printf(*nThe result is %dn*, fun(m); wwjt ();void wwjt ()(FILE *IN,*OUT;int t;i

25、nt o;int c;IN=fopen (in. dat,r);if(IN=NULL)printf(*Read FILE Error*);0UT=fopen (out. dat,w);if(OUT=NULL)(printf(Write FILE Error*)：for(c=l;c=5：c+)fscanf(IN,%d,&t);o=fun(t);fprintf (OUT,%dn”, o);)fclose(IN);fclose(OUT)：答案：-int sum, i;sum =0;for(i=l;i=n;i+)sum=sum+i;return (sum)：第14题题号：507/*【程序设计】功能：编

26、写函数fun求一个字符串的长度,在main函数中输入字符串，并输出其长度。*/include void wwjt ();int fun(char *pl)(/*Program*/* End */ main ()(char *p, a20;int len;p=a;printf(*please input a string:n*);gets(p);len=fun(p);printf(The string* s length is:%dn*, len); wwjt ();void wwjt ()(FILE *IN,*0UT;char *pIN, sin20;int iOUT, i;pIN=sin;I

27、N=fopen(3. IN,1);if(IN=NULL)printf(*Please Verify The Currernt Dir. it May Be Changed);0UT=fopen(3. out,w);if(OUT=NULL)(printf(*Please Verify The Current Dir. it May Be Changed*);)for(i=0;i10;i+)(fscanf (IN,%s”, pIN);iOUT=fun (pIN);fprintf (OUT,%dn, iOUT);)fclose(IN);fclose(OUT);)答案：char *p;int n=0;

28、p=pl：while(*p+)n+;return n;第15题题号:44/*【程序设计】功能：求出二维数组周边元素之和，作为函数值返回。二维数组的值在主函数中赋予。*/ttdefine M 4define N 5#include void wwjt ();int fun(int aMN)/*Prograiri*/mainO(int aMN=1,3,5,7,9,2,4,6,8,10,2,3,4,5,6,4,5,6,7,8;int y;y=fun(a);printf(s=%dn”, y);wwjt();)void wwjt ()(FILE *IN,*OUT;int iINMN, iOUT;int

29、i, j, k;IN=fopen (*14. IN*,*r*);if(IN=NULL)printf(*Please Verify The Currernt Dir. it May Be Changed*);OUT二fopen(14. out,w);if(OUT=NULL)printf(*Please Verify The Current Dir. it May Be Changed*);for(k=0;k10;k+) for(i=0;iM;i+)for(j=0;jN;j+)fscanf (IN,*%d*,&iINij);iOUT=fun(iIN);fprintf(OUT,*%dn*, iOUT

30、);)fclose(IN);fclose(OUT);答案：int s=0;int i, j;for(i=0;iM;i+) s=s+ai0+aiN-l; for(j=l;jN-l;j+)s=s+a0J+aM-lj; return s;第16题题号：364/*【程序设计】功能：求个四位数的各位数字的立方和。*/#include void wwjt ();int fun(int n)(/*Prograiii*/* End */ main()(int k;k=fun(1234);printf(k=%dn, k); wwjt ();)void wwjt ()FILE *IN,*OUT;int ilN,

31、iOUT, i;IN=fopen(in. dat,r); if(IN=NULL)Changed);Changed*);printf(Please Verify The Currernt Dir. it May Be OUT=fopen(out. dat,w)：if(OUT=NULL)(printf(Please Verify The Current Dir. it May Be for(i=0;i0)d=n%10;s+=d*d*d;n/=10;)return s;第17题题号：339/木【程序设计】功能：判断整数X是否是同构数。若是同构数，函数返回1；否则返回0。x的值由主函数从键盘读入，要求

32、不大于1000说明：所谓“同构数”是指这样的数，这个数出现在它的平方数的右边。例如：输入整数5,5的平方数是25,5是25中右侧的数，以5是同构数。*/#include void wwjt ();int fun(int x)/*Prograni*/* End */)main ()int x, y;printf(*nPlease enter a integer numbers:*);scanf &x);if(x100)(printf(*data error!n);exit (0);)y=fun(x);if (y)printf(*%d YESn, x);elseprintf(*%d N0n”,x)

33、;wwjt ();void wwjt ()(FILE *IN,*OUT;int n, t;int o;IN=fopen(*in. dat,r);if(IN=NULL)(printf(Read FILE Error*);0UT=fopen(*out. dat,w);if(OUT二二NULL)printf(*Write FILE Error*);for(t =0; t 5; t+)(fscanf(IN,*%d*,&n);o=fun(n);fprintf (OUT,%dn, o )fclose(IN);fclose(OUT);)答案：int k;k=x*x;if(k%10=x)|(k%100=x)

34、return 1;elsereturn 0;第18题题号：331【程序设计】功能：找出一个大于给定整数且紧随这个整数的素数，并作为函数值返回。*/include #include*conio. hvoid wwjt ();int fun(int n)/*Prograrn*/main ()(int m;printf(*Enter m:);scanf(%d,&m);printf(nThe result is %dn*, fun(m); wwjt ();)void wwjt ()FILE *IN,*OUT;int s ;int t;int o;IN=fopen(*in. dat,r);if(IN=N

35、ULL)(printf(Read FILE Error);OUT二fopen(out. dat,w);if(OUT=NULL)printfCWrite FILE Error);for(s=l;s=5;s+)(fscanf(IN,%d,&t);o=fun(t);fprintf (OUT,%dn”, o);fclose(IN);fclose(OUT);)答案：int i, k;for(i=n+l;i+)for(k=2;ki;k+)if(i%k=O)break;if(k=i) return (i);第19题题号：317卜*【程序设计】功能：求出NXM整型数组的最大元素及其所在的行坐标及列坐标(如果最

36、大元素不唯一，选择位置在最前面的一个)。例如：输入的数组为：12341561218910112求出的最大数为18,行坐标为2,列坐标为1。*/ttdefine N 4#define M 3include void wwjt ();int Row, Col;int fun(int arrayNM)/*Program*/* End */main()int aNM, i, j, max;printf(input a array:);for(i=O;iN;i+)for(j=0;jM;j+)scanf (*%d*,&aij);for(i=0;iN;i+)(for(j=0;jM;j+)printf (*%

37、d*, aij);printf(*n*);max=fun(a);printf (max=%d, row=%d, col=%d”, max, Row, Col);wwjt 0； void wwjt()(FILE *IN,*OUT;int iINNM, iOUT, i, j;IN=fopen(*in. dat,r);if(IN=NULL)(printf(Please Verify The Currernt Dir. it May Be Changed);OUT二fopen(out. dat,w);if(OUT=NULL)(printf(*Please Verify The Current Dir. it May Be Changed*);for(i=0;iN;i+)for(j=0;jM;j+)fscanf (IN,*%d*,&iINij);iOUT=fun