C語言程序設計第四版第八章答案

第八章指針main(){inta,b,c,*p1,*p2,*p3,t;scanf("%d,%d,%d",&a,&b,&c);p1=&a;p2=&b;p3=&c;if(*p1>*p2){t=p1;p1=p2;p2=t;}if(*p1>*p3){t=p1;p1=p3;p3=t;}if(*p2>*p3){t=p2;p2=p3;p3=t;}printf("%d,%d,%d\n",*p1,*p2,*p3);}或main(){inta,b,c,*p1,*p2,*p3,t;scanf("%d,%d,%d",&a,&b,&c);p1=&a;p2=&b;p3=&c;if(a>b){t=*p1;*p1=*p2;*p2=t;}if(a>c){t=*p1;*p1=*p3;*p3=t;}if(b>c){t=*p2;*p2=*p3;*p3=t;}printf("%d,%d,%d\n",a,b,c);}#defineN3#defineMmain(){charstr0[N][M],str1[M],*p,*q;inti,l,m,n;q=str0;for(;p<q+N;p++)gets(p);l=strcmp(q,q+1);if(l>0){strcpy(str1,q);strcpy(q,q+1);strcpy(q+1,str1);}m=strcmp(q,q+2);if(m>0){strcpy(str1,q);strcpy(q,q+2);strcpy(q+2,str1);}n=strcmp(q+1,q+2);if(n>0){strcpy(str1,q);strcpy(q+1,q+2);strcpy(q+2,q+1);}for(p=q;p<q+N;p++)puts(p);}輸入101010個數(shù)。f(x,n)intx[],n;{int*p0,*p1,i,j,t,y;i=j=x[0];p0=p1=x;for(y=0;y<n;y++){if(x[y]>i){i=x[y];p0=&x[y];}elseif(x[y]<j){j=x[y];p1=&x[y];}}t=*p0;*p0=x[n-1];x[n-1]=t;t=*p1;*p1=x[0];x[0]=t;return;}main(){inta[10],u,*r;for(u=0;u<10;u++)scanf("%d",&a[u]);f(a,10);for(u=0,r=a;u<10;u++,r++)printf("%d",a[u]);printf("\n");}nmmm寫一函數(shù)：實現(xiàn)以上功能，在主函數(shù)中輸入n個數(shù)和輸出調(diào)整后的n個數(shù)。#defineN10voidshift(p,x)float*p;intx;{floata[N],*q,*o;inti;o=a;q=p;for(i=0;i<x;i++)*(o+i)=*(q+N-x+i);for(p=p+N-1;p>=q;p--)*p=*(p-x);for(i=0;i<x;i++)*(q+i)=*(o+i);return;}main(){floatshuzhu[N],*u,*v;inth,i;u=v=shuzhu;scanf("%f",&h);for(;u<v+N;u++)scanf("%f",u);shift(v,h);for(u=v;u<v+N;u++)printf("%.2f",*u);printf("\n");}有n人圍成一圈，順序排號。從第1個人開始報數(shù)（從1到3報數(shù)，凡報到3出圈子，問最后留下的是原來的第幾號的那位。#defineN5main(){inti,j,k,a[N+1],*p;for(i=0,p=a;p<=a+N;i++,p++)*p=i;p=a+1;k=N;for(i=0,j=1;k!=1;j++){if(p>(a+N))p=a+1;if(*p!=0)i++;if((i-3)==0){*p=0;i=0;k--;}p++;}for(i=1;i<=N;i++)if(a[i]!=0)printf("Thelastnumberis%d\n",a[i]);}8.7nm制成為另一個字符串。#include"stdio.h"#defineN10main(){chara[N+1],b[N+1],*p,*q;int