清华大学数据结构课后答案完整版(C).doc

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 || m==k) f=1;??else??{????for(i=0;i=k-2;i++) temp[i]=0;????temp[k-1]=1;temp[k]=1; //初始化????sum=1;????j=0;????for(i=k+1;i=m;i++,j++) //求出序列第k至第m个元素的值??????temp[i]=2*sum-temp[j];????f=temp[m];??}??return OK;}//fib分析: k阶斐波那契序列的第m项的值f[m]=f[m-1]+f[m-2]+......+f[m-k]????????=f[m-1]+f[m-2]+......+f[m-k]+f[m-k-1]-f[m-k-1]????????=2*f[m-1]-f[m-k-1]所以上述算法的时间复杂度仅为O(m). 如果采用递归设计,将达到O(k^m).?即使采用暂存中间结果的方法,也将达到O(m^2).??? 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[MAXSIZE];??i=0;??while(result[i].sport!=NULL)??{????switch(result[i].schoolname)????{??????case A:????????score[ 0 ].totalscore+=result[i].score;????????if(result[i].gender==0) score[ 0 ].malescore+=result[i].score;牋牋牋牋else score[ 0 ].femalescore+=result[i].score;牋牋牋牋break;牋牋牋case B:牋牋牋牋score[ 0 ].totalscore+=result[i].score;牋牋牋牋if(result[i].gender==0) score].sc[ 0 ].malescore+=result[i].score;牋牋牋牋else score[ 0 ].femalescore+=result[i].score;牋牋牋牋break;牋牋牋厖牋牋厖牋牋厖牋牋}牋牋score[ 0 ].totalscore+i++；??}??for(i=0;i5;i++)??{????printf(School %d:\n,i);????printf(Total score of male:%d\n,score[i].malescore);????printf(Total score of female:%d\n,score[i].femalescore);????printf(