师范考研语言复习课后答案

1.6a、b、c。{intif(max<b)max=b;printf(“最大数为：“%d”,max{float}考以下问题：（１）c1、c2（２）c1c2ASCIIputcharprintf（３）整形变量与字符变量是否在任何情charc1，c2；int{charc1,c2;}{charc1,c2;printf("c1=%dc2=%d\n",c1,c2);printf("c1=%c}#include<stdio.h>{intif(b<c)printf("max=%d\n",c);elseprintf("max=%d\n",b);elseif(a<c)printf("max=%d\n",c);elseprintf("max=%d\n",a);}#include<stdio.h>{inta,b,c,temp,max;printf3 /*将a和b中的大者存人temp中*/ /*将a和b中的大者与c比较，取最大者*/printf("3个整数的最大数是%d\n”,max);} #include<stdio.h>{int /*x<1{printf("x=%3d,}elseif /*1≤x-10{y=2*x-printf("x=%3d,y=2*x-} /*x≥10{y=3*x-printf("x=%3d,y=3*x-}}分为’B’，70~79C’，60’D’。#includestdio.h>{floatscore;chargrade;printf("\nscanf("%f",&score);}{casecase9:case8:case7:case6:casecasecasecasecasecase0:}printf("成绩是%5.1f，相应的等级是%c。\n",score,grade(score/10)的值进行强制类型转换，得到一个整型值。5321123。 #include<stdio.h>{longintintindiv,ten,hundred,thousand,ten_thousand,if(num>9999)ce=5;elseif(num>999)ce=4;elseif(num>99)ce=3;elseif(num>9)ce=2;elsece=1;printf("ce=%d\n",ce);switch({case5:printf("%d,%d,%d,%d,%d",ten_thousand,thousand,hundred,ten,indiv);printf("\ncase4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);printf("\ncase3:printf("%d,%d,%d",hundred,ten,indiv);printf("\ncase2:printf("\ncase1:break;}}#include<stdio.h>{intprintf4printf("\na=%d,b=%d,c=%d,d=%d\n",a,b,c,d);if(a>b){t=a;a=b;b=t;}if(a>c){t=a;a=c;if(a>d){t=a;a=d;if(b>c){t=a;b=c;if(b>d){t=b;b=d;if(c>d){t=c;c=d;d=t;}printf("%d,%d,%d,}第六 循环控{long{if(m%i==0&&n%i==0)s=i;}if(m>=n)j=m;elsej=n;printf("s=%ld,j=%ld\n",s,j);}{charc;int{if(c>=65&&c<=90||c>=97&&c<=122)i++;elseif(c>=48&&c<=57)j++;elseif(c==32)k++;elsel++;}}求Sn=a+aa+aaa+…+aa…aaa（na）之值，其中a（n=5），n{intn,sum=0,i=1,s=2;}求，（{intn,i=1;longsum=0,s=1;while(i<=n){s=s*i;sum=sum+s;i++;}}求{doublei=1,j=1,k=1,s1=0,s2=0,s3=0,sum;for(;i<=100;i++)s1=s1+i;for(;j<=50;j++)s2=s2+j*j;for(;k<=10;k++)s3=s3+1/k;}153153=1^3+5^3+3^3。{intwhile(x>=100&&x<1000){a=0.01*x;b=10*(0.01*x-a);c=x-100*a-10*b;if(x==(pow(a,3)+pow(b,3)+pow(c,3)))printf("%5d",x);x++;}}610006its factorsare1、2、3{intm,i,j,s;if(m%i==0)s=s+i;{printf("%5ditsfastorsare1",m);for(j=2;j<m;j++)if(m%j==0)printf("%d",j);printf("\n");}}}有一分数序列 求出这个数列的前20项之和{inti=1,n;doublet,x=1,y=2,s,sum=0;while(i<=n)}10{inti,n;doubleh=100,s=100;{h*=0.5;if(i==1)continue;s=2*h+s;}printf("h=%f,s=%f\n",h,s);}上又将剩下的桃子一半，又多吃一个。以后每天早上都吃了前一天剩下的一半零一个。到第10天{inti=1,sum=0;}用迭代法求。求平方根的迭代为： {floatx0,x1,a;}用迭代法求方程在1.5附近的根{doublex,y;x=1.5;printf("x=%.3f\n",x);}用二分法求方程在(-10，10){doublex1,x2,y1,y2;x1=-10;x2=10;}* {inti,j,k;{for(j=0;j<=2-printf("");}printf(""); 第七章{inti,j,a[100];{a[i]=i;printf("%5d",a[i]);}}printf("\n");10{inti,j,a[10],t;}求一个3×3矩阵对角线元和{intprintf("s1=%d,s2=%d\n",s1,s2);{staticinta[5]={1,4,5,6,7};inti,t,b;printf("%d",a[i]);}}{inti,b[10];for(i=9;i>-1;i--打印出以下三角形（要求打印出10行）1 10 ∶∶{staticintm,n,k,b[15][15];{b[m][n]=b[m-1][n-1]+b[m-printf("%-5d",b[m][n]);}printf("\n");}}380{inti,j=0,k=0,l=0,m=0,n=0;charstr0[301],str1[100],str2[100],str3[100];{if(str0[i]>=65&&str0[i]<=90)elseif(str0[i]>=97&&str0[i]<=122)elseif(str0[i]>=48&&str0[i]<=57)l++;elseif(str0[i]==32)m++;elseprintf("DaxieXiaoxieShuziKonggeQita\n");printf("%5d%7d%5d%6d%4d\n",j,k,l,m,n);}**********************＊ *{inti,j,k;chara[5][5];printf("\n");有一行电文译文下面规律译成A- a-B- b-C- c-26i（26-i+1）个字母。非字母字符不变，要求编程序将回原文，并打印出和原文。{inti;charstr1[100],str2[100];elseif(str1[i]>=97&&str1[i]<=122)}strcat{inti,j;charstr1[100],str2[100],str3[201];}S1S2S1>S2，输出一个正数；S1=S2，0；S1<S2，ASCII，’A’与’C’相比，由于’A’<’C’，应输出负数，由于’A’与’C’2，因此应输出"-2"。同理："And"和"Aid"2符比较结果，’n’比’i’5，因此应输出"5"。#include<stdio.h>#include<string.h>{intchars1[100],s2[100];printf("\ninputstring1：");printf("\nInputstring2：");while(s1[i]==s2[i]&&s1[i]!=’\0’)if(s1[i]==’\0’&&s2[i]==’0’)elseresu=s1[i]-s2[i];printf("\nresult：%d\n",resu);}s2s1strcpy #include"stdio.h"{chars1[80],s2[80];inti;printf("Inputs2：");}第八章1.1intm,n;{inti=1,t;t=i;}return(t);}intm,n;{intj;if(m>=n)j=m;elsej=n;returnj;}{intprintf("entertwonumberis:");printf("max=%d,min=%d\n",max,min);}数输入a、b、cfloatyishigen(m,n,k)floatm,n,k;{floatx1=(-x2=(-n-printf("twoshigenisx1=%.3fand}floatdenggen(m,n)floatm,n;{floatx;printf("denggenisx=%.3f\n",x);}floatxugen(m,n,k)float{floatx,y;x=-y=sqrt(-printf("twoxugenisx1=%.3f+%.3fiandx2=%.3f-%.3fi\n",x,y,x,y);}{floata,b,c,q;printf("inputabcis");q=b*b-if(q>0)yishigen(a,b,q);elseif(q==0)denggen(a,b);elsexugen(a,b,q);}8.2intm;{inti=2,t;if(m%i==0&&i<m)break;if(m-i==0)t=1;elset=0;returnm;}{intprintf("entersushuis\n");if(s==1)printf("aissushu\n");elseprintf("sisnotsushu\n");}intzhuangzhi(b)intb[3][3];{inti,j,t;}{inta[3][3];inti,j;printf("%d",a[i][j]);printf("%d",a[i][j]);}{charputs(str0);}charstr1[100];{intcharstr2[100];strcpy(str2,str1);str1[i]=str2[j];}char{strcat(a,b);}{charstr1[100],str2[100];}写一函数，将两个字符串中的元音字母到另一个字符串，然后输出char{inta[i]==69||a[i]==73||a[i]==85){b[j]=a[i];j++;}}{charstr1[100],str2[100];}charf(b)char{inti=0;{printf("");}{inta,u,v,w,t;charc[4];f(c);chartongji(str0,b)charstr0[100];int{int{if(str0[i]>=65&&str0[i]<=90||str0[i]>=97&&str0[i]<=122)b[0]++;elseif(str0[i]>=48&&str0[i]<=57)b[1]++;elseif(str0[i]==32)b[2]++;elseb[3]++;}}{charstr1[100];staticinti,a[4];printf("zimuShuziKonggeQita\n");printf("%-8d",a[i]);printf("\n");char{intstaticcharif(j>=t)}{charstr0[100],longword[15];}intpaixu(x)intx[];{inti,j,t;if(x[i]>x[i+1])}{inty[10];inti;}用迭代法求根。方程为：，系数a，b，c，d由主函数输入。求X在1附近的一个实根。求出doubleqigen(s,t,u,v)ints,t,u,v;{doublex,y;x=1;return}{inta,b,c,d;doublex;}用递归方法求n阶勒让德多项式的值递归floatp(x0,n)intn;floatx0;{floatif(n==0||n==1)if(n==1)y=x0;elseelsey=((2*n-1)*x0*p(x0,n-1)-(n-1)*p(x0,n-2))/n;return(y);}{floatx,y0;inta,i;}intx;charshu[];{inti=strlen(shu)-1,sum=0;{if{if(48<=shu[i]<=57)sum=sum+(shu[i]-48)*pow(16,(i-1));elseif(65<=shu[i]<=90)sum=sum+(shu[i]-55)*pow(16,(i-1));elseif(97<=shu[i]<=102)sum=sum+(shu[i]-87)*pow(16,(i-1));elsex=0;return}{charshufu[100];intif(x)printf("s(D)=%d\n",s);elseprintf("Thenumberisnot}charx[11];unsignedlongm;int{intfor(i=o-1,j=0;i>-1;i--x[i]=(int)((unsignedlong)(m/pow(10,j))%10)+48;}{unsignedlonga,b;intj,i;intfind(x,y,z)intx,y,z;{intif(x%4==0)elset=0;{if(i==2)s=2-t;elses=0;}}{intscanf("%d%d%d",&year,&month,&date);printf("THEDATEISTHE%dth}9--11C（第三版）第九章编译预处理#defineCHANGE(a,b,t)t=a;a=b;b=a{intc,d,s;}#defineQ(a,b)a%b{intscanf("%d%d",&c,&d);}S=(a+b+c)/2，a、b、cS，area，另一个宏用来area。#defineSSS(m,n,k)#defineAQRT(m,n,k)sqrt(SSS(m,n,k)*(SSS(m,n,k)-m)*(SSS(m,n,k)-n)*(SSS(m,n,k)-k)){floatscanf("%f%f%f",&a,&b,&c);printf("s=%.3f}#defineLEAP_YEAR(y)（读者设计的字符串）if(LEAP_YEAR(year))printf("%disaleapyear",year);elseprintf("%disnotaleapyear",year);#defineLEAPYEAR(y)y%4{inty;if(LEAPYEAR(y))printf("%disanotleapyear\n",y);elseprintf("%disalearyear\n",y);}#definePR1(x)#definePR2(x)#definePR3(x)printf("%6.2f\t%6.2f\t%6.2f\n",x,x,x){float}息都放到这个文件内，另编一个程序文件，用命令#include"fornat.h"以确保能使用这些格式。#defineMAX(a,b,c)(a>((b>c)?b:c))?a:((b>c)?b:c){floatx,y,z,t;}9.10成’b’……’z’变成’ａ’其它字符不变）。用命令来控制是否要译成。例如：#defineCHANGE1则输出。若：#defineCHANGE0则不译为，按原码输出#defineCHANGE1{charstr[80],c;inti=0;{#ifif(str[i]==90||str[i]==122)str[i]=str[i]-elseif(str[i]>=65&&str[i]<90||str[i]>=97&&str[i]<122)str[i]=str[i]+1;}puts(str);第十章{int}#defineN#defineM20{charstr0[N][M],str1[M],*p,*q;inti,l,m,n;puts(p);101010int{int*p0,*p1,i,j,t,y; elseif(x[y]<j) return;}{inta[10],u,*r;printf("%d",a[u]);#defineN5{inti,j,k,a[N+1],*p;if((i-printf("Thelastnumberis#defineN10{charintm;}11.3print,5num、name、sore[3]，print#defineN5structstudent{charcharname[8];int{intI,j{printf(“\Inputscoreofstudent%d:\n”,I+1);scanf(“%d”,&stu[i].score[j]);}printf(“\n”);}print(stu);}print(structstudent{intprint(“\n”);求打印出3门课的总平均成绩，以及最高分的学生的数据（包括学号、、3门课成绩）解：#defineN10structstudent{charnum[6]charname[8]intscore[4]floatavr;{intI,j,max,maxi,sum;floataverage;{printf(“\nInputscoresofstudent{printf(“score%d:”,j+1);scanf(“%d”,&stu[i].score[j]);}} if(sum>max){max=sum; maxi=I;}}printf(“NO.namescore1score2score3{printf(“%5s%10s”,stu[i].num,stu[i].name);printf(“%8.2f\n”,stu[i].avr);printf(“Thehighestscoreis:%s,scoretotal:%d.”stu[maxi].nam