实验课题：做这个实验时采用Open Addressing框架，也可加做Separate Chaining以形成比较。

1 构造散列表，把字符串数组中的各项加入到散列表中string MyBirds[13] = { "robin", "sparrow", "hawk", "eagle", "seagull", "bluejay", "owl", "cardinal", "Jakana", "Moa", "Egret", "Penguin", "hawk" };用C表示，可以是char * MyBirds[13] = { "robin", "sparrow", "hawk", "eagle", "seagull", "bluejay", "owl", "cardinal", "Jakana", "Moa", "Egret", "Penguin", "hawk" };为便于观察冲突现象，初始构造散列表时，表的容量不要过大，对Open Addressing，装载因子为0.5左右，对于Separate Chaining，装载因子为1左右即可。

也不要做rehash（应该改源代码的哪里，如何改）。

建议对源代码做些改动、增加一些输出（建议用条件编译控制这些输出），以便于观察冲突的发生和解决；对于Open Addressing，参考代码的冲突解决方案是用的平方探测（quadratic probing），如果用线性探测（linear probing）的策略，应该对函数findPos做什么修改（冲突解决的策略都集中在那里）#include<stdio.h>#include<stdlib.h>#include "hashquad.h"#include<string.h>#define MinTableSize 26typedef unsigned int Index;typedef Index Position;struct HashTbl;typedef struct HashTbl *HashTable;enum KindOfEntry { Legitimate, Empty, Deleted };struct HashEntry{char *Element;enum KindOfEntry Info;};typedef struct HashEntry Cell;struct HashTbl{int TableSize;Cell *TheCells;};static int NextPrime( int N ){int i;if( N % 2 == 0 )N++;for( ; ; N += 2 ){for( i = 3; i * i <= N; i += 2 )if( N % i == 0 )goto ContOuter;return N;ContOuter: ;}}Index Hash( const char *Key, int TableSize ){return *Key % TableSize;}HashTable InitializeTable( int TableSize ){HashTable H;int i;/* 1*/ if( TableSize < MinTableSize ){/* 2*/ printf( "Table size too small" );/* 3*/ return NULL;}/* Allocate table *//* 4*/ H = (struct HashTbl *)malloc( sizeof( struct HashTbl ) );/* 5*/ if( H == NULL )/* 6*/ printf( "Out of space" );/* 7*/ H->TableSize = NextPrime( TableSize );/* Allocate array of Cells *//* 8*/ H->TheCells = (struct HashEntry *)malloc( sizeof( Cell ) * H->TableSize );/* 9*/ if( H->TheCells == NULL )/*10*/ printf( "Out of space" );/*11*/ for( i = 0; i < H->TableSize; i++ ){H->TheCells[i].Element=(char *)malloc(10*sizeof(char));H->TheCells[i].Info=Empty;}/*12*//*13*/ return H;}Position Find( char *Key, HashTable H ){Position CurrentPos;int CollisionNum;/* 1*/ CollisionNum = 0;/* 2*/ CurrentPos = Hash( Key, H->TableSize );//printf("%d\n",CurrentPos);/* 3*/ while( H->TheCells[ CurrentPos ].Info != Empty &&strcmp(H->TheCells[ CurrentPos ].Element,Key)!=0 )/* Probably need strcmp!! */{if (H->TheCells[ CurrentPos ].Element!= NULL)printf("冲突: %s and %s\n", H->TheCells[ CurrentPos ].Element,Key);/* 4*/ CurrentPos += 2 * ++CollisionNum - 1;/* 5*/ if( CurrentPos >= H->TableSize )/* 6*/ CurrentPos -= H->TableSize;}/* 7*/ return CurrentPos;}void Insert( char *Key, HashTable H ){Position Pos;Pos = Find( Key, H );if( H->TheCells[ Pos ].Info != Legitimate ){/* OK to insert here */H->TheCells[ Pos ].Info = Legitimate;strcpy(H->TheCells[ Pos ].Element,Key);/* Probably need strcpy! */}}/*char*Retrieve( Position P, HashTable H ){return H->TheCells[ P ].Element;}*/void DestroyTable( HashTable H ){free( H->TheCells );free( H );}void main(){int i,x,n;char s[10];HashTable H;char * MyBirds[13] = { "robin", "sparrow", "hawk", "eagle", "seagull", "bluejay", "owl", "cardinal", "Jakana", "Moa", "Egret", "Penguin", "hawk" };printf(" \n");printf("原来的MyBirds：\n\n");printf(" 字符串位置\n");for(i=0;i<13;i++)printf("%8s: %2d\n",MyBirds[i],i+1);/*printf(" \n");printf("生成散列表：\n");n=Hash( Key, H->TableSize );for(i=0;i<13;i++)printf("%8s: %2d\n",MyBirds[i],i+1);*/H=InitializeTable( 29 );printf(" \n");printf("生成散列表：\n\n");printf(" 字符串散列值位置\n");for(i=0;i<13;i++)Insert(MyBirds[i] , H );for(i=0;i<29;i++){if(H->TheCells[i].Info!=Empty)printf("%8s: %2d %d\n",H->TheCells[i].Element,x=Hash(H->TheCells[i].Element,29),n=Find( H->TheCells[i].Element,H));}printf("请输入要查找的值：");scanf("%s",s);for(i=0;i<29;i++){if(strcmp(H->TheCells[i].Element,s)==0)break;}if(i<29)printf("查找成功，位置在：%d\n ",Find( s,H));else printf("查找失败\n");DestroyTable( H );}。