C語言期末復(fù)習(xí)試題教學(xué)提綱

1、學(xué)習(xí)資料試卷編號(hào)：10767所屬語言： C 語言試卷方案： TC_150604083824試卷總分： 570 分共有題型： 1 種一、程序設(shè)計(jì)共57題（共計(jì) 570 分）第 1 題 （10.0分）題號(hào) :374難度:易第 2 章/*-【程序設(shè)計(jì)】-功能：編寫函數(shù)實(shí)現(xiàn)兩個(gè)數(shù)據(jù)的交換，在主函數(shù)中輸入任意三個(gè)數(shù)據(jù)，調(diào)用函數(shù)對這三個(gè)數(shù)據(jù)從大到小排序。-*/#include<stdio.h>void wwjt();void swap(int *a,int *b)/*Program*/* End */main()int x,y,z;scanf("%d%d%d",&x

2、,&y,&z);if(x<y)swap(&x,&y);if(x<z)swap(&x,&z);if(y<z)swap(&y,&z);printf("%3d%3d%3d",x,y,z);wwjt();精品文檔學(xué)習(xí)資料void wwjt()FILE *IN,*OUT;int m,n;int i2;IN=fopen("in.dat","r");if(IN=NULL)printf("Read FILE Error");OUT=fopen(&qu

3、ot;out.dat","w");if(OUT=NULL)printf("Write FILE Error");for(n=0;n<3;n+)for(m=0;m<2;m+)fscanf(IN,"%d",&im);swap(&i0,&i1);fprintf(OUT,"%dn",i0);fprintf(OUT,"%dn",i1);fclose(IN);fclose(OUT);答案： -int k;k=*a;*a=*b;*b=k;-第 2 題 （10.0分

4、）題號(hào) :381難度:易第 2 章/*-【程序設(shè)計(jì)】-功能：求一批數(shù)中小于平均值的數(shù)的個(gè)數(shù)。精品文檔學(xué)習(xí)資料-*/#include<stdio.h>void wwjt();int average_num(int a,int n)/*Program*/* End */main()int n,a100,i,num;scanf("%d",&n);for(i=0;i<n;i+)scanf("%d",&ai);num=average_num(a,n);printf("the num is:%dn",num);w

5、wjt();void wwjt()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");精品文檔學(xué)習(xí)資料for(n=0;n<5;n+)fscanf(IN,"%d",&in);o=ave

6、rage_num(i,5);fprintf(OUT,"%dn",o);fclose(IN);fclose(OUT);答案： -int i,sum=0,k=0;double average;for(i=0;i<n;i+)sum=sum+ai;average=sum*1.0/n;for(i=0;i<n;i+)if(average>ai) k+;return(k);-第 3題（10.0 分）題號(hào):406難度:易第 2章/*-【程序設(shè)計(jì)】-功能：編寫函數(shù)fun 求 1000 以內(nèi)所有7 的倍數(shù)之和。-*/#define N 1000#include <std

7、io.h>void wwjt();int fun(int m)/*Program*/* End */精品文檔學(xué)習(xí)資料void main()int sum;sum=fun(7);printf("%d以內(nèi)所有 %d的倍數(shù)之和為：%dn",N,7,sum);wwjt();void wwjt()FILE *IN,*OUT;int n;int i10;int o;OUT=fopen("out.dat","w");if(OUT=NULL)printf("Write FILE Error");o = fun(6);fpri

8、ntf(OUT,"%dn",o);fclose(IN);fclose(OUT);答案： -int s=0,i;for(i=1;i<N;i+)if(i%m=0)s+=i;return s;-第 4題（10.0 分）題號(hào):324難度:易第 2章/*-【程序設(shè)計(jì)】-功能：能計(jì)算從1 開始到n 的自然數(shù)中偶數(shù)的平方的和，n 由鍵盤輸入，并在main() 函數(shù)中輸出。 （ n 是偶數(shù)）精品文檔學(xué)習(xí)資料-*/#include <stdio.h>void wwjt();int fun(int n)/*Program*/* End */main()int m;printf

9、("Enter m: ");scanf("%d", &m);printf("nThe result is %dn", fun(m);wwjt();void wwjt()FILE *IN,*OUT;int t;int o;int c;IN=fopen("in.dat","r");if(IN=NULL)printf("Read FILE Error");OUT=fopen("out.dat","w");if(OUT=NULL)pri

10、ntf("Write FILE Error");for(c=1;c<=5;c+)精品文檔學(xué)習(xí)資料fscanf(IN,"%d",&t);o=fun(t);fprintf(OUT,"%dn",o);fclose(IN);fclose(OUT);答案： -int sum,i;sum =0;for(i=2;i<=n;i=i+2)sum=sum+i*i;return(sum);-第 5題（10.0 分）題號(hào):2難度:中第 1章/*-【程序設(shè)計(jì)】-題目：寫程序求1-3+5- 7+-99+101 的值要求：使用程序中定義的變量-

11、*/#include"stdio.h"void wwjt();/ 函數(shù)功能：求1-3+5- 7+-+n 的值int fun(int n)int i,s=0,f=1;/i定義為循環(huán)變量，s 為 1-3+5- 7+-n 的值/*Program*/* End */return s;main()精品文檔學(xué)習(xí)資料printf("%d",fun(101);wwjt();void wwjt()FILE *IN,*OUT;int i,n;IN=fopen("in.dat","r");if(IN=NULL)printf("

12、Please Verify The Current 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;i<5;i+)fscanf(IN,"%i",&n);fprintf(OUT,"%ldn",fun(n);fclose(IN);fclose(OUT);答案：for (i

13、=1;i<=n;i+=2)s=s+i*f;f=-f;第 6 題 （10.0分）題號(hào) :357難度:易第 2 章/*-【程序設(shè)計(jì)】-精品文檔學(xué)習(xí)資料功能：根據(jù)整型形參m，計(jì)算如下公式的值：y=1/5+1/6 1/7 1/8 1/9+1/10. 1/(m+5)例如：若m=9，則應(yīng)輸出：1.168229-*/#include <stdio.h>void wwjt();double fun(int m)/*Program*/* End */main()int n;printf("Enter n: ");scanf("%d", &n);p

14、rintf("nThe result is %1fn", fun(n);wwjt();void wwjt()FILE *IN,*OUT;int i ;int 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");精品文檔學(xué)習(xí)資料for(i=0;

15、i<5;i+)fscanf(IN,"%d",&t);o=fun(t);fprintf(OUT,"%fn",o);fclose(IN);fclose(OUT);答案： -double y=0;int i;for(i=0; i<=m; i+)y+=1.0/(i+5); return(y);-第 7 題 （10.0分）題號(hào) :392難度:中第 2 章/*-【程序設(shè)計(jì)】-功能：求一批數(shù)中最大值和最小值的積。-*/#define N 30#include "stdlib.h"#include <stdio.h>v

16、oid wwjt();int max_min(int a,int n)/*Program*/* End */精品文檔學(xué)習(xí)資料main()int aN,i,k;for(i=0;i<N;i+)ai=random(51)+10;for(i=0;i<N;i+)printf("%5d",ai);if(i+1)%5=0) printf("n");k=max_min(a,N);printf("the result is:%dn",k);wwjt();void wwjt()FILE *IN,*OUT;int n;int i10;int o

17、;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;n<10;n+)fscanf(IN,"%d",&in);o=max_min(i,10);fprintf(OUT,"%dn",o);fclose(IN);fclo

18、se(OUT);答案： -精品文檔學(xué)習(xí)資料int i,max,min;max=min=a0;for(i=1;i<n;i+)if(ai>max) max=ai;else if(ai<min) min=ai;return(max*min);-第 8 題 （10.0分）題號(hào) :342難度:中第 2 章/*-【程序設(shè)計(jì)】-功能：在鍵盤上輸入一個(gè)3 行 3 列矩陣的各個(gè)元素的值（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()函數(shù)中輸出。-*/#include <stdio.h>void wwjt();void fun(int a33)/*Program*/

19、* End */main()int i,j,s,a33;for(i=0;i<3;i+)for(j=0;j<3;j+)scanf("%d",&aij);s=fun(a);精品文檔學(xué)習(xí)資料printf("Sum=%dn",s);wwjt();void wwjt()FILE *IN,*OUT;int m;int n;int i33;int o;IN=fopen("in.dat","r");if(IN=NULL)printf("Read FILE Error");OUT=fopen(

20、"out.dat","w");if(OUT=NULL)printf("Write FILE Error");for(m=0;m<3;m+)for(n=0;n<3;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;i<3;i+=2)for(j=0;j<3;j+)sum=sum+a

21、ij;return sum;精品文檔學(xué)習(xí)資料-第 9 題 （10.0分）題號(hào) :364難度:中第 2 章/*-【程序設(shè)計(jì)】-功能：求一個(gè)四位數(shù)的各位數(shù)字的立方和。-*/#include <stdio.h>void wwjt();int fun(int n)/*Program*/* End */main()int k;k=fun(1234);printf("k=%dn",k);wwjt();void wwjt()FILE *IN,*OUT;int iIN,iOUT,i;IN=fopen("in.dat","r");if(IN

22、=NULL)printf("Please Verify The Currernt Dir.it May Be Changed");精品文檔學(xué)習(xí)資料OUT=fopen("out.dat","w");if(OUT=NULL)printf("Please Verify The Current Dir. it May Be Changed");for(i=0;i<5;i+)fscanf(IN,"%d",&iIN);iOUT=fun(iIN);fprintf(OUT,"%dn&qu

23、ot;,iOUT);fclose(IN);fclose(OUT);答案： -int d,s=0;while (n>0)d=n%10;s+=d*d*d;n/=10;return s;-第 10 題 （10.0分）題號(hào) :502難度:中第 20章#include <stdio.h>/*-【程序設(shè)計(jì)】-題目：用 while語句求 1 100的累計(jì)和。-*/void wwjt();int fun(int n)/*Program*/* End */精品文檔學(xué)習(xí)資料void main()int sum = 0;sum=fun(100);printf ("sum = %dn&qu

24、ot;, sum);wwjt();void wwjt()FILE *IN,*OUT;int i;int iIN,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 Chang

25、ed");for(i=0;i<5;i+)fscanf(IN,"%d",&iIN);iOUT=fun(iIN);fprintf(OUT,"%dn",iOUT);fclose(IN);fclose(OUT);答案：int i=1,sum=0;while(i<=n)sum=sum+i;精品文檔學(xué)習(xí)資料i+;return sum;第 11題（10.0 分）題號(hào):53難度:中第 1章/*-【程序設(shè)計(jì)】-功能：求一個(gè)四位數(shù)的各位數(shù)字的立方和-*/#include <stdio.h>void wwjt();int fun(in

26、t n)/*Program*/* End */main()int k;k=fun(1234);printf("k=%dn",k);wwjt();void wwjt()FILE *IN,*OUT;int iIN,iOUT,i;IN=fopen("22.IN","r");if(IN=NULL)精品文檔學(xué)習(xí)資料printf("Please Verify The Currernt Dir.it May Be Changed");OUT=fopen("22.out","w");if(O

27、UT=NULL)printf("Please Verify The Current Dir. it May Be Changed");for(i=0;i<10;i+)fscanf(IN,"%d",&iIN);iOUT=fun(iIN);fprintf(OUT,"%dn",iOUT);fclose(IN);fclose(OUT);答案： -int d,k,s=0;while (n>0)d=n%10;s+=d*d*d;n/=10;return s;-第 12題（10.0 分）題號(hào) :409難度 :易第 20章/*-【程

28、序設(shè)計(jì)】-題目：輸入華氏溫度求攝氏溫度。轉(zhuǎn)換公式為c=5/9 （ f-32 ），輸出結(jié)果取兩位小數(shù)。-*/#include <stdio.h>void wwjt();double fun(double m)/*Program*/精品文檔學(xué)習(xí)資料/* End */void main()double c,f;printf("請輸入一個(gè)華氏溫度:");scanf("%f",&f);c=fun(f);printf("攝氏溫度為:%5.2fn",c);wwjt();void wwjt()FILE *IN,*OUT;int i

29、;double iIN,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;i<5;i+)fscanf(IN,"

30、%f",&iIN);iOUT=fun(iIN);fprintf(OUT,"%fn",iOUT);fclose(IN);fclose(OUT);精品文檔學(xué)習(xí)資料答案：float n;n=(5.0/9.0)*(m-32);return n;第 13 題 （10.0 分）題號(hào) :345難度:中第 2 章/*-【程序設(shè)計(jì)】-功能：將兩個(gè)兩位數(shù)的正整數(shù)a、 b 合并形成一個(gè)整數(shù)放在c中。合并的方式是：將a 數(shù)的十位和個(gè)位數(shù)依次放在c 數(shù)的千位和十位上， b 數(shù)的十位和個(gè)位數(shù)依次放在c 數(shù)的個(gè)位和百位上。例如：當(dāng)a 45， b=12。調(diào)用該函數(shù)后，c=4251 。-*

31、/#include <stdio.h>void wwjt();void fun(int a, int b, long *c)/*Program*/* End */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);精品文檔學(xué)習(xí)資料wwjt();void wwjt ( )FILE *rf, *wf ;int i, a,b ;long

32、 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/10*1000+a%10*10+b/10+b%

33、10*100;-第 14題（10.0 分）題號(hào):335難度:中第 2章/*-【程序設(shè)計(jì)】-功能：求一個(gè)n 位自然數(shù)的各位數(shù)字的積。（ n 是小于 10 的自然數(shù)）-*/#include <stdio.h>#include"conio.h"void wwjt();long fun(long n)/*Program*/精品文檔學(xué)習(xí)資料/* End */main()long m;printf("Enter m: ");scanf("%ld", &m);printf("nThe result is %ldn&qu

34、ot;, fun(m);wwjt();void wwjt()FILE *IN,*OUT;int c ;long 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 Error");for(c=1;c<=5;c+)fscanf(IN,"%ld",&t

35、);o=fun(t);fprintf(OUT,"%ldn",o);fclose(IN);fclose(OUT);精品文檔學(xué)習(xí)資料答案： -long d,s=1;while (n>0)d=n%10;s*=d;n/=10;return s;-第 15題（10.0 分）題號(hào):337難度:中第 2章/*-【程序設(shè)計(jì)】-功能：將兩個(gè)兩位數(shù)的正整數(shù)a、 b 合并形成一個(gè)整數(shù)放在c 中。合并的方式是：將 a 數(shù)的十位和個(gè)位數(shù)依次放在 c 數(shù)的百位和個(gè)位上， b 數(shù)的十位和個(gè)位數(shù)依次放在 c 數(shù)的十位和千位上。例如：當(dāng)a 45， b=12。調(diào)用該函數(shù)后，c=2415 。-*/#inc

36、lude <stdio.h>void wwjt();void fun(int a, int b, long *c)/*Program*/* End */main()int a,b; long c;printf("input a, b:");scanf("%d%d", &a, &b);fun(a, b, &c);精品文檔學(xué)習(xí)資料printf("The result is: %ldn", c);wwjt();void wwjt ( )FILE *rf, *wf ;int i, a,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/10*100+a