小浦原的技术专栏

//ShellSort 希尔排序void ShellSort(int A[],int n){ int i,j,gap; int temp; gap=n/2; while(gap>0) { for(i=gap;i<n;i++) { temp=A[i]; j=i-gap; while(j>=0 && temp<A[j]) {

int BTNodeDepth(BTNode *b){ int lchilddep,rchilddep; if(b==NULL) return 0; else { lchilddep=BTNodeDepth(b->lchild); rchilddep=BTNodeDepth(b->rchild); r

BTNode *CreateBTfromPre(char *pre,char *in,int n,int m){ BTNode *s; char *p,*q,*minp; int minpre,minin ,k; if(n<=0) return NULL; minpre=m+1; for(p=in;p<in+n;p++) for(q

void insertNew(BTNode *tree, BTNode *new_node){ int llcnt(0),lrcnt(0),rrcnt(0); BTNode *lch,*rch,*p,*lb; if(tree->lchild==NULL)//insert tree->lchild=new_node; else if(tree->rc

BTNode *CreatBT(char *post,char *in,int n,int m){ BTNode *s; char *p,*q,*maxp; int maxpost,maxin ,k; if(n<=0) return NULL; maxpost=-1; for(p=in;p<in+n;p++) for(q=post;q<post+m;q++) if(*p==

int find_mid( int *a, int *b, int length){ if (length == 1)//这里应该根据中位数的定义取大小 return a[0]<b[0]?a[0]:b[0]; int i = length/2; if (a[i] == b[i]) return a[i]; else if (a[i

我国古代数学名著《孙子算经》中，记载这样一个问题： “今有物不知其数，三三数之剩二，五五数之剩三，七七数之剩二，问物几何。”用现在的话来说就是：“有一批物品，3个3个地数余2个，5个5个地数余3个，7个7个地数余2个，问这批物品最少有多少个?” 这个问题的解题思路,被称为“孙子问题”、“鬼谷算”、“隔墙算”、“韩信点兵”等等.那么，这个问题怎呢？明朝数学家程大位把这一解法编成四句歌诀：　　三人

问题就不描述了，直接上代码： #includeint chess[8][8]={0};int a[8],b[15],c[15];int sum=0; //统计所有摆法void PutQueen(int n){ int col,i,j; for(col=0;col<8;col++

//折半插入排序算法void BiInsertSort(int *Data,int n){ int i,j,low,high,mid,temp; for(i=1;i<n;i++)//先确定第i个元素应插入的位置 { temp=Data[i]; low=0,high=i-1; while(low<=high) { mid=(low+high)/2;

//冒泡排序算法void BubbleSort(int *Data,int n) { int i,j,temp; bool exchange; for(i=n-1,exchange=true;i>0 && exchange;i-- ) { for(j=0,exchange=false;j<i;j++) { if(Data[j+1]<Data[j

//计算树中叶子结点数int LeafNodes(BTNode *b){ int num1,num2; if(b==NULL) return 0; else if(b->lchild==NULL && b->rchild==NULL) return 1; else { num1=LeafNodes(b->lchild); num2=LeafNodes

//计算树中双分支结点数int DBranchNodes(BTNode *b){ int num1,num2,n; if(b==NULL) return 0; else if(b->lchild==NULL || b->rchild==NULL) n=0; else n=1; num1=DBranchNodes(b->lchild); num2=

//计算树中所有结点数int Nodes(BTNode *b){ int num1,num2; if(b==NULL) return 0; else { num1=Nodes(b->lchild); num2=Nodes(b->rchild); return (num1+num2+1); }} 

//树深度算法int BTNodeDepth(BTNode *b){ int lchilddep,rchilddep; if(b==NULL) return 0; else { lchilddep=BTNodeDepth(b->lchild); rchilddep=BTNodeDepth(b->rchild); return (lchilddep>rc

//计算树中分支结点数int BranchNodes(BTNode *b){ int num1,num2,n; if(b==NULL) return 0; else if(b->lchild==NULL && b->rchild==NULL) n=0; else n=1; num1=BranchNodes(b->lchild); num2=Bra

//MergeSort 二路归并排序void Merge(int A[],int low,int mid,int high){ int *A1; int i=low,j=mid+1,k=0; A1=(int *)malloc((high-low+1)*sizeof(int)); while(i<=mid && j<=high) if(A[i]<A[j]) A1

//QuickSort 快速排序void QuickSort(int A[],int low,int high){ int i=low,j=high; int temp; if(low<high) { temp=A[low]; while(i!=j) { while(j>i && A[j]>temp) j--; if(i<j)

//HeapSort 堆排序void Sift(int A[],int low,int high){ int i=low,j=2*i; int temp=A[i]; while(j<=high) { if(j<high && A[j]<A[j+1]) j++; if(temp<A[j]) { A[i]=A[j]; i=j;

void CBTInsert(BTNode *tree, BTNode *new_node){ int llcnt(0),lrcnt(0),rrcnt(0); BTNode *lch,*rch,*p,*lb; if(tree->lchild==NULL)//insert tree->lchild=new_node; else if(tree->rc