C程序设计第四版谭浩强-习题&例题.docx

第7章函数例7.1#includeint main(){ void print_star(); void print_message(); print_star(); print_message(); print_star(); return 0;}void print_star(){ printf("***********************************\n");}void print_message(){ printf("How do you do!\n");}例7.2int max(int x,int y){ int z; z=x>y?x:y; return(z);}#includeint main(){ int max(int x,int y); int a,b,c; printf("please enter two integer number:"); scanf("%d,%d",&a,&b); c=max(a,b); printf("max is %d\n",c); return 0;}例7.3#includeint main(){ int max(float x,float y); float a,b; int c; printf("please enter two integer number:"); scanf("%f,%f",&a,&b); c=max(a,b); printf("max is %d\n",c); return 0;}int max(float x,float y){ float z; z=x>y?x:y; return(z);}例7.4#includeint main(){ float add(float x,float y); float a,b,c; printf("please enter a and b:"); scanf("%f,%f",&a,&b); c=add(a,b); printf("sum is %f\n",c); return 0;}float add(float x,float y){ float z; z=x+y; return(z);}例7.5#includeint main(){ int max4(int a,int b,int c,int d); int a,b,c,d,max; printf("please enter 4 interger number:"); scanf("%d %d %d %d",&a,&b,&c,&d); max=max4(a,b,c,d); printf("max=%d\n",max); return 0;}int max4(int a,int b,int c,int d){ int max2(int a,int b); int m; m=max2(a,b); m=max2(m,c); m=max2(m,d); return(m);}int max2(int a,int b){ if(a>=b) return a; else return b;}#includeint main(){ int max4(int a,int b,int c,int d); int a,b,c,d,max; printf("please enter 4 interger number:"); scanf("%d %d %d %d",&a,&b,&c,&d); max=max4(a,b,c,d); printf("max=%d\n",max); return 0;}int max4(int a,int b,int c,int d){ int max2(int a,int b); return(max2(max2(max2(a,b),c),d));}int max2(int a,int b){ return(a>b?a:b);}例7.6#includeint main(){ int age(int n); printf("NO.5,age:%d\n",age(5)); return 0;}int age(int n){ int c; if(n==1) c=10; else c=age(n-1)+2; return(c);}例7.7递归措施求n！#includeint main(){ int fac(int n); int n,y; printf("input a interger number:"); scanf("%d",&n); y=fac(n); printf("%d!=%d\n",n,y); return 0;}int fac(int n){ int f; if(n<0) printf("n<0,data error!"); else if(n==0||n==1) f=1; else f=fac(n-1)*n; return(f);}例7.8汉诺塔#includeint main(){ void hanoi(int n,char one,char two,char three); int m; printf("input the number of diskes:"); scanf("%d",&m); printf("The step to move %d diskes:\n",m); hanoi(m,'A','B','C');}void hanoi(int n,char one,char two,char three)//将N-1个盘子从一座移到另一座。

{ void move(char x,char y);//用到这个函数，提前声明 if(n==1) move(one,three);//最后一步1-3 else { hanoi((n-1),one,three,two);//小和尚工作n-1个盘子借用3由1搬到2 move(one,three);//老和尚工作最大盘子由1搬到3 hanoi((n-1),two,one,three);//小和尚工作n-1个盘子借用2由1搬到3，完毕工作 }}void move(char x,char y)//将一种盘子从一座移到另一座{ printf("%c-->%c\n",x,y);}例7.9#includeint main(){ int max(int x,int y); int a[10],m,n,i; printf("enter 10 interger numbers:"); for(i=0;i<10;i++) scanf("%d",&a[i]); printf("\n"); for(i=1,m=a[0],n=0;i<10;i++) { if (max(m,a[i])>m) { m=max(m,a[i]); n=i; } } printf("The largest number is %d\nit is the %dth number.\n",m,n+1);}int max(int x,int y){ return(x>y?x:y);}例7.10#includeint main(){ float average(float array[10]); float score[10],aver; int i; printf("enter 10 scores:\n"); for(i=0;i<10;i++) scanf("%f",&score[i]); printf("\n"); aver=average(score); printf("average score is %5.2f\n",aver); return 0;}float average(float array[10]){ int i; float aver,sum=array[0]; for (i=1;i<10;i++) sum=sum+array[i]; aver=sum/10; return(aver);}例7.11#includeint main(){ float average(float array[10],int n); float score1[5]={98.5,97,91.5,60,55}; float score2[10]={67.5,89.5,99,69.5,77,89.5,76.5,54,60,99.5}; printf("The average of class A is %6.2f\n",average(score1,5)); printf("The average of class B is %6.2f\n",average(score2,10)); return 0;}float average(float array[10],int n){ int i; float aver,sum=array[0]; for (i=1;iint main(){ void sort(int array[],int n); int a[10],i; printf("enter array:\n"); for(i=0;i<10;i++) scanf("%d",&a[i]); sort(a,10); printf("The sorted array:\n"); for(i=0;i<10;i++) printf("%d ",a[i]); printf("\n"); return 0;}void sort(int array[],int n){ int i,j,k,t; for(i=0;iint main(){ int max_value(int array[][4]); int a[3][4]={{1,3,5,7},{2,4,6,8},{15,17,34,12}}; printf("Max valu。