程序设计与训练练习答案

不编入1C略第2DBDC,D,CC,DBBB1,1(1)2,(2)1,(3) , 五、已知“inta=5,b=25,x=5;1．a+b%5*(a+b)%2/5

六、2．35．0,6.a,A,第..9.10. (1)(3)(4)(1)

00(1)a=%d,b=%d,

c1=’%c’or%d(ASCII)c2=’%c’or%d12341234.0%

T

01,0,21,324intmain(){floatreturn}intmain(){inta,b;printf("%d%d%d%f",a+b,a-b,a/b,sqrt(a*a+b*b);return0;}#definePI3.14intmain(){floatr,h;printf("%f%f",2*PI*r*h+2*PI*r*r,PI*r*r*h);return0;}编写一个程序，先将值'x'和76.1分别到不同的变量中，然后输出这两个值intmain(){chara='x';doubleb=76.1;printf("%c%.1f",a,b);return0;}数输入前三个值，第4个值使用赋值语句赋值。intmain(){floatz;inta;printf("%c%c%E%d\n",x,y,z,a);return0;}intmain(){inta;return0;}celsius5(fahrenheit9intmain(){floatfahrenheit;return0;}第4DB,A,CDDCAAB(1)m=1a<b&&011y%4==0&&y%100||75. 2<x&&x<3||x<-1a=2,2int{intx=0,y=2,z=3;switch(x){case0:switch(y==2)case1:printf("*1");break;case2:printf("%");}case1:switch(z)case case2:printf("*2");break;default:printf("#");}}return}main()charc1='B',c2='h';switch(c1){case'A':case'B':switch(c2)case'a': case }case'C':switch(c2)case'g':case'h':}default:printf("&\n");}return}?编写一个交互式程序,if语句用于计算正方形的面积（area=side2）或直1/2×base×eightintmain(){inta;if(a){floatbase,height;}elsefloatside;}return}4-4风速 类 弱 强 大 72以 飓 intmain(){inta;if(a<25)printf("弱风elseif(a<=38)printf("强风\n");elseif(a<=54)printf("大风\n");elseif(a<=72)printf("狂风\n");elseprintf("飓风\n");return}在变量watts中（单位是瓦特4-5所示。4-5功期望亮度（流明功亮度（流明lumens赋值为-1intmain(){intlumens,watts;{{case15:case25:case40:case60:case75:{lumens=1000;break;}case100:{lumens=1675;break;}default:lumens=-1;}}return}intmain(){floata,b,c;))elseif((a*a+b*b)==c*c||(a*a+c*c)==b*b||(b*b+c*c)==a*aprintf("直角三角形\n");elseif(a+b>c&&a+c>b&&b+c>a)printf("一般三角形\n");elseprintf("不能构成三角形\n");return0;} (5xy (yx (0xintmain(){printf("X的值\n");floatx;if(x>-5&&x<0)printf("%f\n",x);elseif(x==0)printf("%f\n",x-1);elseif(x>0&&x<10)printf("%f\n",x+1);elseprintf("所输的值不在考虑范围内");return0;}intmain(){printf("X的值\n");floatx;intflag=3;if(-5<x&&x<0)flag=0;elseif(x==0)flag=1;elseif(x>0&&x<10)flag=2;switch(flag){casecase2:{printf("%f\n",x+1);break;}}return}第5ACACDCAADDB2.3.-4.do-while5.1- x%i==01.25811 5.6.20a=16,01122438456intmain(){inti;{printf("%d",i);intf=pow(2,i);}return}intmain(){intsum=0,i;{}return0;}数,并确定要经过多少年人口数才会超过30000。intmain(){ints=9870,i;if(s>30000){printf("%d\n",i);return}}1+2!+3!+...+20!的和,并输出计算结果。int{intsum=0,s=1,i;for(i=1;i<=20;i++){}return0;}intmain(){intf1=1,f2=1,flag=0,i;printf("%d",f1);if(flag%5==0)printf("\n");printf("%d",f2);}return}intmain(){inti,j;for(i=1;i<=9;i++){printf("%d*%d=%2d",j,i,i*j);}return}#include<stdio.h>intmain(){intfor(i=0;i<5;{or(j=0;j<=i;j++)}return}#include<stdio.h>intmain(){intfor(a=1;a<10;{for(b=1;b<10;}return#include<stdio.h>#include<math.h>intmain(){inti,jfor(i=-3;i<=3;i+//ifor(k=1;k<=abs(i);k++)//k控制空格printf("for(j=1;j<=7-2*abs(i);j+//j*printf("*");}return}#include<stdio.h>//对称intmain(){int{if(abs(i)+abs(j)<=3printf("}return}

。试编写程序计算611161111…+1 n2{intdoubleeps,pie=0;while(1.0/n>{}}6996.7.12，o10.5

12813aThisisCprogram,输出#include<stdio.h>intmain(){inta[10]={8,-inti,t,max_index,min_index;}}}return}intmain(){

{{}{}

c[i]=a[p--printf("%d",c[i]);return}aa为2,5,2,4,20,0,4,-5,-5,16,2,-4,20a为5,-4,0,2,4,5,16,20#include<stdio.h>intmain(){inti,j,k,x,n;}}}return0;}#include<stdio.h>#include<stdlib.h>#defineN13intmain()inti,j,k,x,n,p,t;for(i=0;i<n;i++)for(p=0;p<=n-2;{for(i=p+1,j=p+1;i<=n-1;}for(i=0;i<n;i++)for(i=0;i<n-1;{for(j=i+1;j<n;j++)if(a[j]<a[p])p=j;if(p!=i)}}for(i=0;i<n;i++)return0;}算法，采用折半查找算法确定待元素的位置，在数组中元素5和-5。#include<stdio.h>intmain(){intx,a[12]={-9,-6,-intleft,right,mid,i;printf("输入要的数}

return}#include<stdio.h>#defineN8int{inta[N]={-intprintf("输入要删除的数while(pos<n&&a[pos]!=x){}

n--printf("没有值为%d的元素return0;}int{

n--i--}}return}a、ba、bcinta[]={-9,-6,-3,0,2,3,7,11,14,20};intb[]={-10,-8,-6,-3,0,5,7,11,14,19},c中的元素有{-6,-3,0,7,11,14}。#include<stdio.h>//本题可以理为求集合a与集合b的交集cintmain(){inta[]={-9,-6,-intb[]={-10,-8,-6,-3,0,5,7,11,14,19};//abintt=0;inti,j;for({for({{}}}printf("%d",c[i]);return0;} A- A∪B-#defineU6#defineV{

cho1(intu[],intv[],intinti,j,w_l;{}}{

returnwl;intu[U]={12,0,-5,4,6,-

v[V]={13,-5,9,-8,4},return}#defineU6#defineV#defineW{

cho2(intu[],intv[],intint{--}}}{

returnwl;intu[U]={12,0,-5,4,6,-

v[V]={13,-5,9,-8,4},return}#include#defineU#defineV#defineWintcho3(intu[],int{

intintu_l=6,v_l=5,w_l;{--}}}{

returnw_l;intu[U]={12,0,-5,4,6,-

v[V]={13,-5,9,-8,4},return}#include<stdio.h>intinti,j,len,temp;intflag;}}return}#defineN5int{floata[N][N];inti,j,k;{{if(i==j)a[i][j]*=-elseif(i<j)a[i][j]-=3;elsea[i][j]+=3;}printf("%5.2f",a[i][k]);}return}intmain(){intinti,j;}return}素存放到4*3的二维数组中。intmain(){floata[3][4],b[12],c[4][3],min;intk,i,j; }{}{{printf("%4.2f",c[i][j]);}}return}n[3][4]中每行最小的元素移到副对角线上，最小元素和该行其他元素的相intmain(){intinti,j,k;//if(n[i][j]<min[i])}intt=n[i][0];for(k=1;k<4;k++){n[i][k-}p--}{printf("%2d",n[i][j]);}#includestdio.h>intmain(){

for(i=0;i<4;i++){b[8]+=a[}while(b[j]<x&&j>=0){j--}}return0;}等于（,=~#include<stdio.h>intmain(){inti,j;}return}#defineN8int{inta[N][N];floats=0;inti,j;{if(i==0||i==N-1)s+=1.0*a[i][j]/(N*4-4);elseif(j==0||j==N-1)s+=1.0*a[i][j]/(N*4-4);}return0;}12345122212345331#defineN5int{intfor(i=0;i<N;i++){for}if(i==j||i+j==n-elseif(i<j&&i+j<n-1)b[i][j]=2;elseif(i>j&&i+j<n-1)b[i][j]=3;elseif(i>j&&i+j>n-1)b[i][j]=4;elseb[i][j]=5;for(i=0;i<N;i++){for}return} AAAAAAAAAA#defineN4intinti,j,k;{{for(k=0;k<N-i-1;k++)printf("");for(k=0;k<N-i-1;k++)printf("}}return}#include<stdio.h>#defineN4voids[][N]){ints[j][i-//s[i-intintn[N][N],i,j;}return}#defineN4intinti,j;printf("");for(j=-i;j<=i;j++)}return}intmain(){charss[100];inti=0;{}return0;} 数是5。intmain(){

intword=0,flag=0;charch;printf(“\n”);elseif(flag==1)}if(flag==1)word++;return0;}intmain(){chara[80]={"CfhIJSVwXyz"};charc='m';inti=0,j,len=strlen(a);return0;}intmain(){chara[80]={"CfhIJSVwXyz"};inti;return0;}intcharchart[80];return}#include<stdio.h>intchars[80]="ThisisCprogram!";inti,len,j;return0;}substr的子串(区分大小写)。例如：str[]=“Thisprogram”,sub[]=“pro”,str中包含sub，sub就是str的子串。#include<stdio.h>#include<string.h>intmain(){

ischarstr[100],sub[80];inti,j,k,flag=0;{if(flag)break;{')//{}}return0;}str[]=“ThisisCprogram”既包含“pro”sub[]=“ramThis”substr的intmain(){chars[]="ThisisCinti,j,k,n,ls,lt;{printf("%s不是%s的子串{

{;}printf("%s\t不是%s的子串}}

printf("%s\t是%s的子串return}第7 初始字符串为Theoriginal移动后的字符串为tTheoriginalTheresultis:47.5338.5,-3,5,8,8,8,8,-3,-i=2j=3i=3j=3 a=3b=210. ThisisC元音字母有114.编写一个int link(chars[],chars1[],chars2[])函数将字符串s2连接到s1后存入s3中。函数返回s3字符串长度。#defineMAX1#defineMAX2#defineMAX3200chars3[MAX3];intlink(chars1[],chars2[],chars3[]){inti=0,j=0;}while(s2[j]!='\0'){s3[}return}intcharinti,length;return}11an返回这些数的个数。intfun(){inti;{}return}intintlength=fun();inti;printf("%d",a[i]);return}int(int,int)，最小公倍数函数intlcm(int,int)。int(intx,y){inti,m=x<y?if(x%i==0&&y%i==0)returni;}intlcm(intx,y){inti,m=x<y?if(i%x==0&&

1 12 123...floatfun(int{floats=0;int{}return}int{intn;return0;}2222 2 2 22222 2 2 2 2...intmain(){doublepi,item,next;return0;}n856sum=763。intfun(int{ints=0,i;{if(n%i==0)}return}int{inta;return0;}所谓回文数是指各位数字左右对称，例如121、676、94249等。满足上述条件的数如m=11,m*m=121,m*m*m=1331intalue(longm)实现此功能，如果10。intalue(longm){longp=m,s=0;{}{}{}}

returnreturnintfor(long{}return}intfun(intk,intn){inti=0;doublep=0;{p+=(int)pow(10,i)*(k%10);}return}int{inta,b;return0;}n,intp)，将p进制数(p<10n转换为十进制数。并比较这两个函数的异同点。 n,p){}return}

main(){ntm,n,to;scanf("%d%d",&m,&to);//m=12345to=8printf("10进制的%d=%d进制%d\n",m,to,n);return0; n,p){} intm,n,from;scanf("%d%d",&m,&from);//m=30071from=8printf("%d进制的%d=10进制%d\n",from,m,n);return0;}

则 165非回文

#include<stdio.h>#include<stdlib.h>intreverse(intx){ n=x}return}intispalin(int x==}intx){inty;return}main(){i x,printf("%d\n",x);printf("%d\n",y);return0;}voidselect(intr[],intn){int){}}voidbubble(intr[],intn){inti,j,temp;}}}int earch(intr[],intx){intlow=0,high=n-1,mid;if(x==r[mid])returnmid;}

return-}intdelelem(inta[],intn)a的n个元素中的重复数据，只保留一个。函数返回数组中剩余元素的个数。a的初始数据为:2,5,2,4,20,0,4,-5,-5,16,2,-4,20。经delelem函数处理后数组a的元素为:-5,-4,0,2,4,5,16,20。#include<stdio.h>#include<stdlib.h>intdelelem(inta[],intn){inti,k;}return}voidsort(inta[],intn){int}}}intreturn}#include<stdio.h>#includevoidreverse(chars[],intleft,intright){if(left<right){chart=s[left];}}intchars[40]={"CProgram!"};intleft,right;return0;} P(x) ((2n (x)(n

nnnP4(1.5)。floatp(intn,float{if(n==0)returnif(n==1)return}int{intn;floatx;return0;}#include<stdio.h>#defineN8voidBubbleSort(intr[],int{inti,t;{{}}}int{return}#include<stdio.h>#defineN8voidSelectSort(intr[],int{inti,k,t;{{}}}int{return0;}#includestdio.h>#defineN8intBinarySearch(intr[],intx,intleft,int{intmid=left+(right-left)/2;return-1;returnmid;}int{intprintf("输入要查找的数(Ctrl+Z结束{printf("%d是第%d个元素printf("输入要查找的数(Ctrl+Z结束}return}第 1．A，2．C， “w”ch,fp

{

FILE*fpr,*fpw;charch;

if((fpr=fopen("clock.avi","r"))==NULL){\n!\n");}if((fpw=fopen("d:clock.avi","w"))==NULL){\nn");}while(!feof(fpr)){ch=fgetc(fpr}return0;}

intfloata,b,c,d,term1,term2;inti;FILE*fpr,*fpw;}if((fpw=fopen("output.txt","w"))==NULL){printf("\noutput.txt}for(i=0;i<3;i++){%f",&a,&b,&c);d=b