第九章9．1定義一個結構體變量 （包括年、月、日）。計算該日在本年中是第幾天，

注意閏年問題。解：Struct{intyear;intmonth;intday;}date;main(){intdays;printf( “Inputyear,month,day: ”);scanf(“%d,%D,%d”,&date.year,&date.month,&date.day);switch(date.month){case1:days=date.day; break;case2:days=date.day+31; break;case3:days=date.day+59; break;case4:days=date.day+90; break;case5:days=date.day+120; break;case6:days=date.day+31; break;case7:days=date.day+181; break;case8:days=date.day+212; break;case9:days=date.day+243; break;case10:days=date.day+273; break;case11:days=date.day+304; break;case12:days=date.day+334; break;}if((date.year%4==0&&date.year%100!=0||date.year%400==0)&&date.month>=3)days+=1;printf(

n%d/%d“isthe%dthdayin%d.

”,date.month,data.day,days,date,year);}9.2寫一個函數

days,實現上面的計算。由主函數將年、月、日傳遞給

days

函數，計算后將日數傳回主函數輸出。解：structy_m_d{intyear:intmonth;intday;}date;intdays(structy_m_ddate1){intsum;switch(data.month){case1:sum=date1.day;case2:sum=date1.day+31;case3:sum=date1.day+59;case4:sum=date1.day+90;case5:sum=date1.day+120;case6:sum=date1.day+151;

break;break;break;break;break;break;case7:sum=date1.day+181; break;case8:sum=date1.day+212; break;case9:sum=date1.day+243; breakcase10:sum=date1.day+243; breakcase11:sum=date1.day+243; breakcase12:sum=date1.day+243; break}};9.3編寫一個函數 print,打印一個學生的成績數，該數組中有錄包括num、name、sore[3]，用主函數輸入這些記錄，用解：

5個學生的數據記錄，每個記print 函數輸出這些記錄。#defineN5structstudent{charnum[6];charname[8];intscore[4];}stu[N];main(){intI,j;for(I=0;I<N;I++){printf(

“

n”,I+1);printf(

“no.:

”);scanf(

“%s”,stu[i].num);printf(

“name:”);scanf( “%s”,stu[i].name);for(j=0;j<3;j++){printf( “score%d:”j+1);scanf( “%d”,&stu[i].score[j]);}printf( n“”);}print(stu);}print(structstudentstu[6]){intI,j;printf( “%5s%10s”,stu[i].num,stu[i].name);for(j=0;j<3;j++)printf(

“%9d”,stu[i].score[j]);print(

“n”);}9.4在上題的基礎上，編寫一個函數

input,

用來輸入

5個學生的數據記錄。解：#defineN5structstudent{charnum[6];charname[8];intscore[4]}stu[N];input(structstudentstu[]){intI,j;for(I=0;I<N;I++){printf( “inputscoresofstudent n”,I+1);printf( “NO.:”);scanf( “%s”,stu[i].num);printf( “name: ”);scanf( “%s”,stu[i].name);for(j=0;j<3;j++){printf( “score%d:”,j++);scanf( “%d”,&stu[i].score[j]);}}printf( n“”);}}9.5有10個學生，每個學生的數據包括學號、姓名、 3門課的成績，從鍵盤輸入 10個學生的數據，要求打印出 3門課的總平均成績，以及最高分的學生的數據（包括學號、姓名、 3門課成績）解：#defineN10structstudent{charnum[6]charname[8]intscore[4]floatavr;}stu[N];main(){intI,j,max,maxi,sum;floataverage;for(I=0;I<N;I++){printf(

“

n”,I+1);printf( “NO.:”);scanf( “%s”,stu[i].num);printf( “name”);scanf( “%s”,stu[i].name);for(j=0;j<3;j++){printf( “score%d: ”,j+1);scanf( “%d”,&stu[i].score[j]);}}average=0;max=0;maxi=0;for(i=0;i<3;i++){sum=0;for(j=0;j<3;j++)sum+=stu[i].score[j];stu[i].avr=sum/3.0;average+=stu[i].avr;if(sum>max){max=sum;maxi=I;}}average/=N;printf( “NO.namescore1score2score3averagen”);for(I=0;I<N;I++){printf( “%5s%10s”,stu[i].num,stu[i].name);for(j=0;j<3;j++)printf(printf(

“%9d”,stu[i].score[j]);“%8.n2f”,stu[i].avr);}printf(printf(

“average=%6.2fn”,average);“Thehighestscoreis:%s,scoretotal:%d.

”stu[maxi].name,max);}9.6編寫一個函數new,對n個字符開辟連續的存儲空間，此函數應返回一個指針，指向字符串開始的空間。New(n)表示分配n個字節的內存空間。解：new函數如下：#defineNULL0#defineNEWSIZE1000charnewbuf[NEWSIZE];char*newp=newbuf;char*new(intn){if(newp+n<=newbuf+NEWSIZE){newp=newp+n;return(newp-n);}elsereturn(NULL);}9.7寫一函數 free，將上題用 new函數占用的空間釋放。 Free(p)表示將p指向的單元以后的內存段釋放。解：#defineNullo#defineNEWSIZE1000charnewbuf[NEWSIZE];char*newp=newbuf;free(char*p){if((p>=newbuf)&&(p<newbuf+NEWSIZE))newp=p;}9.8已有a、b亮光鏈表，每個鏈表中的結點包括學好、成績。要求把兩個鏈表合并，按學號升序排列。解：#include<stdio.h>#defineNULL0#defineLENsizeof(structstudent)strutstudent{longnum;intscor;structstudent*next};structstudentlistA,listB;intn,sum=0;main(){structstudent*creat(void);structstudent*insert(structstudent*,structstudent*);voidprint(structstudent*);stuctstudent*ahead,*bhead,*abh;printf( “ n”);ahead=creat();sum=sum+|n;abh=insert(ahead,bhead);print(abh);}structstudent*creat(void){structstudent*p1,*p2,*head;n=0;p1=p2=(structstudent*)malloc(LEN);printf( “inputnumber&scoresofstudent:n”);printf( “ifnumberIs0,stopinputing.n”);scanf( “%ld,%d”,&p1->num,&p1->score);head=NULL;while(p1->num!=0){n=n+1;if(n==1)head=p1;elsep2->next=p1;p2=p1;p1=(structstudent*)malloc(LEN);scanf( “%ld,,%d”,&p1->num,&p1->score);}p2->next=NULL;return(head);}structstudent*insert(structstudent*ah,structstudent*bh){structstudent*pa1,*pa2,*pb1,*pb2;pa2=pa1=ah;pb2=pb1=bh;do{while((pb1->num>pa1->num)&&(pa1->next!=NULL)){pa2=pa1;pa1=pa1->next;}if(pb->num<=pa1->num){if(ah=pa1)ah=pb1;elsepa2->next=pb1;pb1=pb1->next;pb2->next=pa1;pa2=pb2;pb2=pb1;}}while((pa1->next!=NULL)||(pa1==NULL&&pb1!=NULL));if((pb1->num>pa1->num)&&(pa1->next==NULl))ap1->next=pb1;return(ah);}voidprint(structstudent*head){structstudent*p;printf( “%ld%dn”,p->num,p->score);p=p->next;while(p!=NULL);}9.913個人圍成一圈，從第

1個人開始順序報號

1、2、3。凡報到“

3”者退出圈子。找出最后留在圈子中的人原來的序號。解：#defineN13structperson{intnumber;intnextop;}link[N+1];main(){intI,count,h;for(I=1;I<=N;I++){if(I==N)link[i].nextp=1;elselink[i].nextp=I+1;link[i].number=I;}printf( n“”);count=0;h=N;printf( “sequencethatperson2leavethecircle:n”);while(count<N-1){I=0;while(I!=3){h=link[h].nextp;if(link[h].number)I++;}printf( “%4d”,link[h].number);link[h].number=0;count++;}printf( \nThe“lastoneis”);for(I=1;ii<=N;I++)if(link[i].number)printf( “%3d”,lin[i].number);}9.10有兩個鏈表

a和

b,設結點中包含學號、姓名。從

1鏈表中刪去與

b鏈表中有相同學號的那些結點。解：#defineLA4#defineLB5#defineNULL0structstudent{charnump[6];charname[8];structstudent*next;}A[LA],b[LB];main(){structstudenta[LA]={{structstudentb[LB]={{

“101”,”Wang”},{“103”,”Zhang”},{

“102”,”LI”},{“104”,”Ma”},{

“105”,”zhang”},{ “106”,”“105”,”Chen”},{ “107”,”Gu{“108”,”Lui”}};intI,j;structstudent*p,*p1,*p2,*pt,*head1,*head2;head1=a;head2=b;printf( “listan”:);for(p1=head1,i=1;p1<a=LA;i++){p=p1;p1->next=a+I;printf( “%8s%8sn”,p1->num,p1->name);p1=p1->next;}p->next=NULL;printf( “ n”);for(p2=head2,I=1;p2<b+LB;I++){p=p2;p2->next=b+I;printf(“%8s%8sn”,p2->num,p2->name);p2=pa->next;}p->next=NULL;printf( n“”);p1=head1;while(p1!=NULL){p2=head2;while(p2!=NULL&&strcmp(p1->num,p2->num)!=0)p2=p2->next;if(strcmp(p1->num,p2->num==0))if(p1==head1)head1=p1->next;elsep->next=p1->next;p=p1;p1=p1->next;}p1=hedad1;printf{ “ n”};while(p1!=NULL){printf( “%7s%7sn”,p1->num,p1->name);p1=p1->next;}}9.11建立一個鏈表，每個結點包括：學號、姓名、性別、年齡。輸入一個年齡，如果鏈表中的結點所包含的年齡等于此年齡，則將此結點刪去。解：#defineNULL0#defineLENsizeof(structstudent)structstudent{charnum[6];charname[8];charsex[2];intage;stuctstudent*next;}stu[10];main(){structstudent*p,*pt,*head;intI,length,iage,flag=1;intfind=0;while(flag==1){printf(“inputlengthoflist （<10）:”);scanf( “%d”,&length);if(length<10)flag=0;}for(I=0;I<lenth;I++){p=(structstudent*)malloc(LEN);if(I==0)head=pt=p;elsept->next=p;pt=p;ptintf( “NO:”);scanf( “%s”->num);,pprntf( “name:”);scanf( “%s”->name);,pprintf( “sex:”);scanf( “%s”->sex);,pprintf( “age:”);scanf( “%s”->age);,p}p->next=NULL;p=head;printf( “ n”);while(p!=NULL){printf( “%4s%8s%6s%6dn”,p->num,p->name,p->sex,p->age);p=p->next;}printf( “Inputage: ”);scanf( “%d”,&iage);pt=head;p=pt;if(pt->age==iage){p=pt->next;head=pt=p;find=1;}elsept=pt->next;while(pt!=NULL){if(pt->age==iage){p->next=pt->next;find=1;}elsep=pt;pt=pt->next;}if(!find)printf( “Notfound%d. ”,iage);p=head;printf( /r/