C语言上机模拟试题1

C语言机试模拟试题一、改错题：(30分)1、求一维数组a中的值为偶数的元素之和。#include<stdio.h>sum(intarr[],intn){inti,s;s=0;for(i=0;i<n;i++)if(arr[i]%2==0)/************found************/s=s+i;改为：s=s+arr[i];return(s);}voidmain(){inta[10]={10,4,2,7,3,12,5,34,5,9},i,s;/************found************/s=sum(a,2);改为：s=sum(a,10);printf("Theresultis:%d\n",s);}2、求一维数组a中的值为奇数的元素之和。#include<conio.h>#include<stdio.h>intsum(intb[],intn){inti,s=0;for(i=0;i<n;i++)if(b[i]%2==1)/************found************/s=s+b[i]改为：s=s+b[i];return(s);}voidmain(){inta[12]={10,4,2,7,3,12,5,34,5,9,21,19},n;/************found************/n=sum(a,2);改为：n=sum(a,12);printf("Theresultis:%d\n",n);}3、求二维数组a中的最大值和最小值。#include<stdio.h>voidmain(){inta[3][3]={4,4,34,37,3,12,5,6,5},i,j,max,min;//clrscr();max=min=a[0][0];for(i=0;i<3;i++)/************found************/for(j=1;j<3;j++)改为:for(j=0;j<3;j++){if(max<a[i][j])max=a[i][j];/************found************/if(min<a[i][j])改为：if(min>a[i][j])min=a[i][j];}printf("Themaxis:%d\n",max);printf("Theminis:%d\n",min);}4、求一维数组a中的最大元素及其下标。#include<stdio.h>voidmain(){inta[10]={1,4,2,7,3,12,5,34,5,9},i,max,pos;//clrscr();max=a[0];pos=0;for(i=1;i<10;i++)/************found************/if(max>a[i])改为：if(max<a[i]){max=a[i];/************found************/i=pos;改为：pos=i;}printf("Themaxis:%d,posis:%d\n",max,pos);}5、猴子吃桃的问题，求第一天的桃子数。#include<stdlib.h>#include<math.h>voidmain(){inti,j,x,p;//clrscr();x=1;/************found************/j=12;改为：j=11;while(j>=1){p=(x+1)*2;/************found************/p=x;改为：x=p;j--;}printf("totalis%d\n",p);6、求一维数组a中的值为奇数的元素的平均值。#include<stdio.h>doubleaverage(intarr[],intn){intk=0,i;doubles;s=0;for(i=0;i<n;i++)/************found************/if(arr[i]%2=1)改为：if(arr[i]%2==1){s=s+arr[i];k++;}return(s/k);}voidmain(){inta[12]={10,4,2,7,3,12,5,34,5,9,21,18};doubles;//clrscr();/************found************/s=average(a[12],12);改为：s=average(a,12);printf("Theresultis:%.2f\n",s);}二、填空题：(30分)1、求一维数组a中的素数的个数：2、程序的功能是：在第一个循环中从键盘上给a数组的前M(M<100)个数组元素依次赋值，在第二个循环中使a数组前M个元素中的值对称折叠，在最后一个循环中输出折叠后的a数组的前M个元素。程序如下：#defineM5voidmain(){inti,a[100],t;//clrscr();for(i=0;i<M;i++)/**************found************/(1)___-填写的语句为：scanf(“%d”:,&a[i]);/**************found************/for(i=0;—⑵—;i++)-填写的语句为：i<M/2a[M-i-1]=a[i];for(i=0;i<M;i++)printf("%5d",a[i]);printf("\n");}3、计算正整数num的各位上的数字之和。#include<stdio.h>voidmain(){intnum,k;//clrscr();/************found************/—⑴—；填写的语句为：k=0;printf("Pleaseenteranumber:");scanf("%d",&num);do/************found************/k=(2);填写的语句为：k=k+num%10;num/=10;}while(num);printf("\n%d\n",k);}4、求cman=m!/n!*(m-n)!之值，例如：m=12,n=2时，cmanis66#include<math.h>#include<stdio.h>longintfun(intx){longints=1;inti;for(i=1;i<=x;i++)s*=i;/************found************/⑴；填写的语句为：returns;}voidmain(){longintm=12,n=2,s1,s2,s3;//clrscr();s1=fun(m);s2=fun(n);s3=fun(m-n);/************found************/printf("Cmnis:%ld”,(2));填写的语句为：s1/(s2*s3)}5、输出Fabonacci数列1,1,2,3,5……前20项，要求每位输出5个数。#include"stdio.h"voidmain(){longintf[20]={1,1};inti;//clrscr();/**************found************/—⑴—;填写的语句：intj=0;for(i=2;i<20;i++){f[i]=f[i-1]+f[i-2];/**************found************/if(—(2)—)printf("\n");填写的语句：j++%5==0printf("%8ld",f[i]);}printf("\n");}6、求两个正整数[m,n]之间所有既能被3整除也能被7整除的整数之和。#include<math.h>#include<stdio.h>main(){intm,n,i,t;longints=0;//clrscr();scanf("%d,%d",&m,&n);if(m>n){t=m;m=n;n=t;}/************found************/for(⑴;i++)填写的语句：i=m;i<=n;if(i%3==0&&i%7==0)s+=i;/************found************/printf("Sumis:⑵\n",s);填写的语句：％ld}7、从键盘输入一个由数字组成的字符串(字符串长度小于8)，将该字符串转换成一个十进制数。例如：从键盘输入12345,则程序运行的结果应当为：n=12345#include"stdio.h"#include"string.h"voidmain(){chars[10];inti;longintn;//clrscr();/**************found************/—⑴—；填写的语句：n=0scanf("%s",s);for(i=0;i<strlen(s);i++)/**************found************/n=n*10+(2);填写的语句：i+1printf("n=%ld\n",n);}8、从键盘输入一个正整数n,计算并输出n的阶乘。#include<math.h>#include<conio.h>#include<stdio.h>longfac(intn){inti;longt=1;for(i=2;i<=n;i++)t=t*i;return(t);}voidmain(){intn;longints;//clrscr();printf("Enteranintegernumber:");/************found************/⑴;填写的语句：scanf("%ld",&n);s=fac(n);/************found************/printf("Theresultis:(2)\n",s);填写的语句：％ld}三、编程题：(40分)1、按下面的公式求s的近似值，直到最后一项值的绝对值小于x为止，x的值由键盘输入,结果保留4位小数。S=1-1/3+1/5-1/9+…..+(-1)A(n-1)1/2n-1#include"stdio.h"#include"math.h"#include"stdlib.hdoublefun(doublex)/***********begin***********/inty,i=1;doublea,sum=0.0;do{if(i%2==0)y=-1;elsey=1;a=y*1.0/(2*i-1);sum=sum+a;i++;}while(fabs(a)>=x);returnsum;/***********end***********/}voidmain(){doublex,sum;printf("Enterx:\n");scanf("%lf",&x);sum=fun(x);printf("sum=%.4f\n",sum);NONO();}NONO(){FILE*fr,*fw;inti;doublex;fr=fopen("D:\\exam\\03200310\\PROGIN.DAT","r");fw=fopen("D:\\exam\\03200310\\PROGOUT.DAT","w");for(i=1;i<=5;i++){fscanf(fr,"%lf",&x);fprintf(fw,"s=%.4f\n",fun(x));}fclose(fr);fclose(fw);}2、将一个4位的整数m拆开，然后按相反的顺序输出，每输出一位数其后留2个空格。如m=1234,输出的结果应为m=1234,4321.程序如下：#include"stdio.h"#include"math.h"#include"stdlib.hvoidmain(){intm=1234,a,b,c,d;/***********begin***********/a=m%10;b=m/10%10;c=m/100%10;d=m/1000;/************end************/printf("m=%d,%d%d%d%d\n",m,a,b,c,d);NONO(m,a,b,c,d);}NONO(x,a,b,c,d)intx,a,b,c,d;{FILE*f;f=fopen("D:\\exam\\03200302\\PROGOUT.DAT","w");fprintf(f,"x=%d:%3d%3d%3d%3d####\n",x,a,b,c,d);fclose(f);}3、按下面的公式，求S的近似值。S=1+1/2*2+1/3*3+...+1/n*n#include"stdio.h"#include"math.h"#include"stdlib.hdoublefun(intn)/***********begin***********/inti;doublep,sum=0.0;for(i=1;i<=n;i++){p=i*i;sum=sum+1.0/p;}returnsum;/************end************/}voidmain(){doubles;intn;//clrscr();printf("Enters:\n");scanf("%d",&n);s=fun(n);printf("s=%.3f\n",s);NONO();}NONO(){FILE*fr,*fw;inti,m;fr=fopen("D:\\exam\\03200306\\PROGIN.DAT","r");fw=fopen("D:\\exam\\03200306\\PROGOUT.DAT","w");for(i=1;i<=5;i++){fscanf(fr,"%d",&m);fprintf(fw,"s=%.3f####\n",fun(m));}fclose(fr);fclose(fw);}a)求一个三数m的各位上的数字之和。m由键盘输入。#include<stdio.h>intfun(intm){/***********begin***********/ints=0;while(m!=0){s=s+m%10;m=m/10;}returns;/************end************/}voidmain(){intm,s;//clrscr();printf("Enterm:\n");scanf("%d",&m);s=fun(m);printf("Resultis:%d\n",s);NONO();}NONO(){FILE*fr,*fw;inti,j,k,m;fr=fopen("D:\\exam\\03200308\\PROGIN.DAT","r");fw=fopen("D:\\exam\\03200308\\PROGOUT.DAT","w");for(i=1;i<=5;i++){fscanf(fr,"%d",&m);fprintf(fw,"Sum=%d\n",fun(m));}fclose(fr);fclose(fw);}b)求两个正整数m和n的最大公约数，将求出的最大公数存入变量t中。#include"stdio.h"#include"math.h"#include"stdlib.hvoidmain(){intm=76,n=40,t,i;/***********begin***