数据结构一、基础实例

基础的数据结构实例：
1.1
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(x<y) x<->y; //<->为表示交换的双目运算符,以下同
  if(y<z) y<->z;
  if(x<y) x<->y; //冒泡排序
  printf("%d %d %d",x,y,z);
}//print_descending
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1.2

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(k<2||m<0) return ERROR;
  if(m<k-1) f=0;
  else if (m==k-1) f=1;
  else
  {
    for(i=0;i<=k-2;i++) temp[i]=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;j<i;j++) sum+=temp[j];
      temp[i]=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).
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1.3
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[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;
        else score[ 0 ].femalescore+=result[i].score;
        break;
      case 'B':
        score .totalscore+=result[i].score;
        if(result[i].gender==0) score .malescore+=result[i].score;
        else score .femalescore+=result[i].score;
        break;
      ……    ……    ……
    }
    i++；
  }
  for(i=0;i<5;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("Total score of all:%d\n\n",score[i].totalscore);
  }
}//summary
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1.4
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++)
  {
  a[i-1]=last*2*i;
   if((a[i-1]/last)!=(2*i)) reurn OVERFLOW;
   last=a[i-1];
   return OK;
  }
}//algo119
分析:当某一项的结果超过了maxint时,它除以前面一项的商会发生异常.
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1.5
void polyvalue()
{
  float ad;
  float *p=a;
  printf("Input number of terms:");
  scanf("%d",&n);
  printf("Input the %d coefficients from a0 to a%d:\n",n,n);
  for(i=0;i<=n;i++) scanf("%f",p++);
  printf("Input value of x:");
  scanf("%f",&x);
  p=a;xp=1;sum=0; //xp用于存放x的i次方
  for(i=0;i<=n;i++)
  {
    sum+=xp*(*p++);
    xp*=x;
  }
  printf("Value is:%f",sum);
}//polyvalue