#include <iostream>
#include "string.h"
#include "fstream"
#define NULL 0
unsigned int key
unsigned int key2
int *p
struct node //建节点
{
char name[8],address[20]
char num[11]
node * next
}
typedef node* pnode
typedef node* mingzi
node **phone
node **nam
node *a
using namespace std//使用名称空间
void hash(char num[11]) //哈希函数
{
int i = 3
key=(int)num[2]
while(num[i]!=NULL)
{
key+=(int)num[i]
i++
}
key=key%20
}
void hash2(char name[8]) //哈希函数
{
int i = 1
key2=(int)name[0]
while(name[i]!=NULL)
{
key2+=(int)name[i]
i++
}
key2=key2%20
}
node* input() //输入节点
{
node *temp
temp = new node
temp->next=NULL
cout<<"输入姓名："<<endl
cin>>temp->name
cout<<"输入地址："<<endl
cin>>temp->address
cout<<"输入电话："<<endl
cin>>temp->num
return temp
}
int apend() //添加节点
{
node *newphone
node *newname
newphone=input()
newname=newphone
newphone->next=NULL
newname->next=NULL
hash(newphone->num)
hash2(newname->name)
newphone->next = phone[key]->next
phone[key]->next=newphone
newname->next = nam[key2]->next
nam[key2]->next=newname
return 0
}
void create() //新建节点
{
int i
phone=new pnode[20]
for(i=0i<20i++)
{
phone[i]=new node
phone[i]->next=NULL
}
}
void create2() //新建节点
{
int i
nam=new mingzi[20]
for(i=0i<20i++)
{
nam[i]=new node
nam[i]->next=NULL
}
}
void list() //显示列表
{
int i
node *p
for(i=0i<20i++)
{
p=phone[i]->next
while(p)
{
cout<<p->name<<'_'<<p->address<<'_'<<p->num<<endl
p=p->next
}
}
}
void list2() //显示列表
{
int i
node *p
for(i=0i<20i++)
{
p=nam[i]->next
while(p)
{
cout<<p->name<<'_'<<p->address<<'_'<<p->num<<endl
p=p->next
}
}
}
void find(char num[11]) //查找用户信息
{
hash(num)
node *q=phone[key]->next
while(q!= NULL)
{
if(strcmp(num,q->num)==0)
break
q=q->next
}
if(q)
cout<<q->name<<"_" <<q->address<<"_"<<q->num<<endl
else cout<<"无此记录"<<endl
}
void find2(char name[8]) //查找用户信息
{
hash2(name)
node *q=nam[key2]->next
while(q!= NULL)
{
if(strcmp(name,q->name)==0)
break
q=q->next
}
if(q)
cout<<q->name<<"_" <<q->address<<"_"<<q->num<<endl
else cout<<"无此记录"<<endl
}
void save() //保存用户信息
{
int i
node *p
for(i=0i<20i++)
{
p=phone[i]->next
while(p)
{
fstream iiout("out.txt", ios::out)
iiout<<p->name<<"_"<<p->address<<"_"<<p->num<<endl
p=p->next
}
}
}
void menu() //菜单
{
cout<<"0.添加记录"<<endl
cout<<"3.查找记录"<<endl
cout<<"2.姓名散列"<<endl
cout<<"4.号码散列"<<endl
cout<<"5.清空记录"<<endl
cout<<"6.保存记录"<<endl
cout<<"7.退出系统"<<endl
}
int main()
{
char num[11]
char name[8]
create()
create2()
int sel
while(1)
{
menu()
cin>>sel
if(sel==3)
{ cout<<"9号码查询，8姓名查询"<<endl
int b
cin>>b
if(b==9)
{ cout<<"请输入电话号码:"<<endl
cin >>num
cout<<"输出查找的信息:"<<endl
find(num)
}
else
{ cout<<"请输入姓名:"<<endl
cin >>name
cout<<"输出查找的信息:"<<endl
find2(name)}
}
if(sel==2)
{ cout<<"姓名散列结果:"<<endl
list2()
}
if(sel==0)
{ cout<<"请输入要添加的内容:"<<endl
apend()
}
if(sel==4)
{ cout<<"号码散列结果:"<<endl
list()
}
if(sel==5)
{ cout<<"列表已清空:"<<endl
create()
create2()
}
if(sel==6)
{ cout<<"通信录已保存:"<<endl
save()
}
if(sel==7) return 0
}
return 0
}
应该是这个意思：第一次冲突就是散列的位置+1，这次发生冲突了就继续第二次
第二次用的是平方取中，55^2= 3025，当然第二次冲突的RH2就是02了，答案(2)
#include<stdio.h>int main(){
char s[256]
char *p
unsigned long long int h = 0
scanf("%s", s)
for(p=s *p p++){
h = h*31 + *p
}
printf("%llu", h)
}
