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C语言程序设计(谭浩强)第四版-课后答案解析

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for (j=1;j<=i;j++)

printf(\ for (k=0;k<5;k++) printf(\ }

printf(\ return 0; }

6-12a-c #include <> int main() { int j,n;

char ch[80],tran[80];

printf(\ gets(ch);

printf(\ :%s\ j=0;

while (ch[j]!='\\0')

{ if ((ch[j]>='A') && (ch[j]<='Z')) tran[j]=155-ch[j];

else if ((ch[j]>='a') && (ch[j]<='z')) tran[j]=219-ch[j]; else

tran[j]=ch[j]; j++; } n=j;

printf(\ for (j=0;j

6-12b

#include <> int main() {int j,n;

char ch[80];

printf(\ gets(ch);

printf(\ j=0;

while (ch[j]!='\\0')

{ if ((ch[j]>='A') && (ch[j]<='Z')) ch[j]=155-ch[j];

else if ((ch[j]>='a') && (ch[j]<='z')) ch[j]=219-ch[j]; else

ch[j]=ch[j]; j++; } n=j;

printf(\ for (j=0;j

#include <> int main()

{ char s1[80],s2[40]; int i=0,j=0;

printf(\ scanf(\

printf(\ scanf(\ while (s1[i]!='\\0') i++;

while(s2[j]!='\\0') s1[i++]=s2[j++]; s1[i]='\\0';

printf(\ return 0; } 6-14

#include <> int main() { int i,resu;

char s1[100],s2[100]; printf(\ gets(s1);

printf(\ gets(s2); i=0;

while ((s1[i]==s2[i]) && (s1[i]!='\\0'))i++; if (s1[i]=='\\0' && s2[i]=='\\0') resu=0; else

resu=s1[i]-s2[i];

printf(\ return 0; }

6-15

#include <> #include <> int main()

{ char s1[80],s2[80]; int i;

printf(\ scanf(\

for (i=0;i<=strlen(s2);i++) s1[i]=s2[i];

printf(\ return 0; }

第7章用函数实现模块化程序设计7-1-1

#include <> int main()

{int hcf(int,int); int lcd(int,int,int); int u,v,h,l;

scanf(\ h=hcf(u,v); printf(\ l=lcd(u,v,h); printf(\ return 0;

218页】 【第 }

int hcf(int u,int v) {int t,r; if (v>u)

{t=u;u=v;v=t;} while ((r=u%v)!=0) {u=v; v=r;} return(v); }

int lcd(int u,int v,int h) {

return(u*v/h); }

7-1-2

#include <> int Hcf,Lcd; int main()

{void hcf(int,int); void lcd(int,int); int u,v;

scanf(\ hcf(u,v); lcd(u,v);

printf(\ printf(\ return 0; }

void hcf(int u,int v) {int t,r; if (v>u)

{t=u;u=v;v=t;} while ((r=u%v)!=0) {u=v; v=r; } Hcf=v; }

void lcd(int u,int v) {

Lcd=u*v/Hcf; }

7-2

#include <> #include <>

float x1,x2,disc,p,q; int main()

{void greater_than_zero(float,float); void equal_to_zero(float,float); void smaller_than_zero(float,float); float a,b,c;

printf(\

scanf(\

printf(\ disc=b*b-4*a*c; printf(\ if (disc>0) {

greater_than_zero(a,b);

printf(\ }

else if (disc==0)

{equal_to_zero(a,b);

printf(\ } else

{smaller_than_zero(a,b);

printf(\ }

return 0; }

void greater_than_zero(float a,float b) {x1=(-b+sqrt(disc))/(2*a); x2=(-b-sqrt(disc))/(2*a); }

void equal_to_zero(float a,float b) {

x1=x2=(-b)/(2*a);

C语言程序设计(谭浩强)第四版-课后答案解析

for(j=1;j<=i;j++)printf(\for(k=0;k<5;k++)printf(\}printf(\return0;}6-12a-c#includeintmain(){intj,n;charch
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