【精品】清华大学数据结构习题集(C版)答案.doc

清华大学严蔚敏数据结构习题集（C版）答案 清华大学严蔚敏数据结构习题集（C版）答案 第一章 绪论 1.16 void print_descending(int x,int y,int z)//按从大到小顺序输出三个数{? scanf(%d,%d,%d,x,y,z);? if(xy) x-y; //-为表示交换的双目运算符,以下同? if(yz) y-z;? if(xy) x-y; //冒泡排序? printf(%d %d %d,x,y,z);}//print_descending 1.17 Status fib(int k,int m,int f)//求k阶斐波那契序列的第m项的值f{? int tempd;? if(k2||m0) return ERROR;? if(mk-1) f=0;? else if (m==k-1) f=1;? else? {??? for(i=0;i=k-2;i++) temp=0;??? temp[k-1]=1; //初始化??? for(i=k;i=m;i++) //求出序列第k至第m个元素的值??? {????? sum=0;????? for(j=i-k;ji;j++) sum+=temp[j];????? temp=sum;??? }??? f=temp[m];? }? return OK;}//fib分析:通过保存已经计算出来的结果,此方法的时间复杂度仅为O(m^2).如果采用递归编程(大多数人都会首先想到递归方法),则时间复杂度将高达O(k^m). 1.18 typedef struct{??????????????????? char *sport;??????????????????? enum{male,female} gender;??????????????????? char schoolname; //校名为A,B,C,D或E??????????????????? char *result;??????????????????? int score;????????????????? } resulttype; typedef struct{??????????????????? int malescore;??????????????????? int femalescore;??????????????????? int totalscore;????????????????? } scoretype; void summary(resulttype result[ ])//求各校的男女总分和团体总分,假设结果已经储存在result[ ]数组中{? scoretype score;? i=0;? while(result.sport!=NULL)? {??? switch(result.schoolname)??? {????? case A:??????? score[ 0 ].totalscore+=result.score;??????? if(result.gender==0) score[ 0 ].malescore+=result.score;??????? else score[ 0 ].femalescore+=result.score;??????? break;????? case B:??????? score.totalscore+=result.score;??????? if(result.gender==0) score.malescore+=result.score;??????? else score.femalescore+=result.score;??????? break;????? ……??? ……??? ……??? }??? i++；? }? for(i=0;i5;i++)? {??? printf(School %d:\n,i);??? printf(Total score of male:%d\n,score.malescore);??? printf(Total score of female:%d\n,score.femalescore);??? printf(Total score of all:%d\n\n,score.totalscore);? }}//summary 1.19 Status algo119(int a[ARRSIZE])//求i!*2^i序列的值且不超过maxint{? last=1;? for(i=1;i=ARRSIZE;i++)