【正文】 )。for(p=number。}void output(int *number){ int *p。number[9]=*max。if(max==number)max=min。number[0]=*min。else if(*pmin=p。for(p=number+1。}void max_min_value(int *number){int *max,*min,*p,temp。iscanf(“%d”,amp。printf(“input 10 numbers:”)。output(number)。input(number)。void output(int *)。寫3個(gè)函數(shù)：1輸入10個(gè)數(shù)；2 進(jìn)行處理；3 輸出10個(gè)數(shù) include void main(){void input(int *)。strcpy(p2,p)。strcpy(p,p1)。printf(“%sn%sn%sn”,str1,str2,str3)。if(strcmp(str2,str3)0)swap(str2,str3)。if(strcmp(str1,str2)0)swap(str1,str2)。gets(str2)。printf(“input three line:n”)。include include void main(){ void swap(char *,char *)。*p2=p。p=*p1。printf(“%d,%d,%dn”,a,b,c)。if(*p1*p3)swap(p1,p3)。scanf(“%d,%d,%d”,p1,p2,p3)。p3=amp。p2=amp。p1=amp。int a,b,c。} if(sign==1 || loca==1)printf(“%d not been ”,n)。}else if(nbott=mid1。printf(“NO.%d , his name is %”,n,name[loca])。(topmid=(bott+top)/2。while((sign==1)amp。sign=1。bott=N1。}void search(int n,int num[],char name[N][8]){ int top,bott,mid,loca,sign。for(i=0。strcpy(name[min],temp2)。strcpy(name[i],name[min])。strcpy(temp2,name[i])。jif(num[min]num[j])min=j。imin=i。char temp2[8]。gets(name[i])。printf(“input name: ”)。scanf(“%d”,amp。for(i=0。)flag=0。||c==39。if(c==39。getchar()。search(number,num,name)。scanf(“%d”,amp。sort(num,name)。char name[N][8]。void search(int ,int [],char name[][8])。include include define N 10void main(){ void input(int [],char name[][8])。}714 寫幾個(gè)函數(shù)： 輸入10個(gè)職工的姓名和職工號(hào) 按職工號(hào)由小到大的順序排序，姓名順序也隨之調(diào)整。sumxn+=a_stu[i]。for(i=0。sumx=。}float s_var(void){ int i。c=j+1。jif(score[i][j]high){high=score[i][j]。for(i=0。int i,j。a_cour[j]=s/(float)N。for(i=0。for(j=0。} }void aver_cour(void){ int i,j。js+=score[i][j]。for(i=0。} }void aver_stu(void){ int i,j。jscanf(“%f”,amp。iprintf(“ninput score of student%2d:n”,i+1)。}void input_stu(void){ int i,j。cour2cour3 cour4cour5printf(“highest:%NO.%2dcourse %2dn”,h,r,c)。printf(“n”)。for(j=0。printf(“%”,a_stu[i])。for(j=0。for(i=0。aver_cour()。input_stu()。void aver_stu(void)。float highest()。float h。int r,c。include define N 10 define M 5 float score[N][M]。找出所有50個(gè)分?jǐn)?shù)中最高的分?jǐn)?shù)所對(duì)應(yīng)的學(xué)生和課程。1 計(jì)算每個(gè)學(xué)生平均分。39。j++。U39。u39。O39。o39。I39。i39。E39。e39。A39。a39。39。for(i=0,j=0。printf(“The vowel letters are:%sn”,c)。gets(str)。char str[80],c[80]。}77 寫一個(gè)函數(shù)，將字符竄中元音字母復(fù)制到另一個(gè)字符竄，然后輸出。string[i+j]=39。39。for(j=0。39。for(i=0。printf(“nThe new string is %sn”,s)。scanf(“%s”,s2)。scanf(“%s”,s1)。char s1[100],s2[100],s[100]。} }76 寫一個(gè)函數(shù)，將兩個(gè)字符竄連接。str[i]=str[j1]。for(i=0,j=strlen(str)。}void inverse(char str[]){ char t。inverse(str)。printf(“input string:”)。include include char str[100]。array[j][i]=t。j{t=array[i][j]。for(i=0。printf(“n”)。ifor(j=0。}printf(“convert array:n”)。jprintf(“%5d”,array[i][j])。for(i=0。array[i][j])。ifor(j=0。printf(“input array:n”)。void main(){void convert(int array[][3])。}74 寫一個(gè)函數(shù)，使給定的一個(gè)3x3的二維整形數(shù)組轉(zhuǎn)置，即行列轉(zhuǎn)換。iif(n%i==0)flag=0。}int prime(int n){ int flag=1,i。if(prime(n))printf(“n %d is a ”,n)。scanf(“%d”,amp。int n。}73 寫一個(gè)判斷素?cái)?shù)的函數(shù)，在主函數(shù)輸入一個(gè)整數(shù)，輸出是否素?cái)?shù)的信息。}void smaller_than_zero(float a,float b){ p=b/(2*a)。x2=(bsqrt(disc))/(2*a)。printf(“x1=%f+%fitx2=%f%fin”,p,q,p,q)。printf(“x1=%fttx2=%fn”,x1,x2)。printf(“x1=%fttx2=%fn”,x1,x2)。printf(“root:n”)。printf(“equation: %*x*x+%*x+%=0n”,a,b,c)。b,amp。scanf(“%f,%f,%f”,amp。float a,b,c。void equal_to_zero(float,float)。include include float x1,x2,disc,p,q。} 72 求方程ax*x+bx+c=0的根，用3個(gè)函數(shù)分別求當(dāng)b*b4ac大于0，等于0、小于0時(shí)的根并輸出結(jié)果。} return(v)。} while((r=u%v)!=0){u=v。u=v。}int hcf(int u,int v){ int t,r。l=lcd(u,v,h)。h=hcf(u,v)。u,amp。int u,v,h,l。第一篇：C語言課后習(xí)題解答第七章 函數(shù)調(diào)用 71 寫兩個(gè)函數(shù)，分別求兩個(gè)整數(shù)的最大公約數(shù)和最小公倍數(shù)，include void main(){ int hcf(int,int)。int lcd(int,int,int)。scanf(“%d,%d”,amp。v)。printf(“=%dn”,h)。printf(“=%dn”,l)。if(vu){t=u。v=t。v=r。}int lcd(int u,int v,int h){ return(u*v/h)。從主函數(shù)輸入abc的值。void main(){ void greater_than_zero(float,float)。void smaller_than_zero(float,float)。printf(“input a,b,c:”)。a,amp。c)。disc=b*b4*a*c。if(disc0){greater_than_zero(a,b)。} else if(disc==0){equal_to_zero(a,b)。} else {smaller_than_zero(a,b)。} }void greater_than_zero(float a,float b){ x1=(b+sqrt(disc))/(2*a)。} void equal_to_zero(float a,float b){ x1=x2=(b)/(2*a)。q=sqrt(disc)/(2*a)。include void main(){ int prime(int)。printf(“ninput an integer:”)。n)。elseprintf(“n %d is not a ”,n)。for(i=2。return(flag)。include define N 3 int array[N][N]。int i,j。for(i=0。jscanf(“%d”,amp。printf(“noriginal array :n”)。ifor(j=0。printf(“n”)。for(i=0。jprintf(“%5d”,array[i][j])。} }void convert(int array[][3]){ int i,j,t。ifor(j=i+1。array[i][j]=array[j][i]。} }75 寫一個(gè)函數(shù)，是輸入的字符串按反序存放，在主函數(shù)中輸入和輸出字符竄。void main(){ void inverse(char str[])。scanf(“%s”,str)。printf(“inverse string:%sn”,str)。int i,j。it=str[i]。str[j1]=t。include void main(){ void concatenate(char string1[],char string2[],char string[])。printf(“input string1:”)。printf(“input string2:”)。concatenate(s1,s2,s)。}void concatenate(char string1[],char string2[],char string[]){ int i,j。string1[i]!=39。i++)string[i]=string1[i]。string2[j]!=39。j++)string[i+j]=string2[j]。39。include void main(){ void cpy(char [],char [])。printf(“input string:”)。cpy(str,c)。}void cpy(char s[],char c[]){int i,j。s[i]!=39。i++)if(s[i]==39。||s[i]==39。||s[i]==39。||s[i]==39。||s[i]==39。||s[i]==39。||s[i]==39。||s[i]==39。||s[i]==39。||s[i]==39。){c[j]=s[i]。} c[j]=39。}713 輸入10個(gè)學(xué)生5門課的成績，分別用函數(shù)實(shí)現(xiàn)下列功能。2 計(jì)算每門課的平均分。4 計(jì)算平均方差。float a_stu[N],a_cour[M]。void main(){int i,j。float s_var(void)。void input_stu(void)。void aver_cour(void)。aver_stu()。printf(“n avern”)。iprintf(“n NO %2d ”,i+1)。jprintf(“%”,score[i][j])。}printf(“naverage:”)。jprintf(“%”,a_cour[j])。h=highest()。printf(“variance %”,s_var())。for(i=0。for(j=0。score[i][j])。float s。ifor(j=0,s=0。a_stu[i]=s/。float s。js=0。is+=score[i][j]。} }float highest(){ float high。high=score[0][0]。ifor(j=0。r=i+1。} return(high)。float sumx,sumxn。sumxn=。isumx+=a_stu[i]*a_stu[i]。} return(sumx/N(sumxn/N)*(sumxn/N))。要求輸入一個(gè)職工號(hào)，用折半查找法找出該職工的姓名，從主函數(shù)輸入要查找的職工號(hào)，輸出該職工姓名。void sort(int [],char name[][8])。