11-散列3 QQ帐户的申请与登陆 (25 分)
实现QQ新帐户申请和老帐户登陆的简化版功能。最大挑战是:据说现在的QQ号码已经有10位数了。
输入格式:
输入首先给出一个正整数N(≤10^5),随后给出N行指令。每行指令的格式为:“命令符(空格)QQ号码(空格)密码”。其中命令符为“N”(代表New)时表示要新申请一个QQ号,后面是新帐户的号码和密码;命令符为“L”(代表Login)时表示是老帐户登陆,后面是登陆信息。QQ号码为一个不超过10位、但大于1000(据说QQ老总的号码是1001)的整数。密码为不小于6位、不超过16位、且不包含空格的字符串。
输出格式:
针对每条指令,给出相应的信息:
1)若新申请帐户成功,则输出“New: OK”;
2)若新申请的号码已经存在,则输出“ERROR: Exist”;
3)若老帐户登陆成功,则输出“Login: OK”;
4)若老帐户QQ号码不存在,则输出“ERROR: Not Exist”;
5)若老帐户密码错误,则输出“ERROR: Wrong PW”。
输入样例:
5
L 1234567890 myQQ@qq.com
N 1234567890 myQQ@qq.com
N 1234567890 myQQ@qq.com
L 1234567890 myQQ@qq
L 1234567890 myQQ@qq.com
输出样例:
ERROR: Not Exist
New: OK
ERROR: Exist
ERROR: Wrong PW
Login: OK
#include<iostream> #include<string> #include<map> using namespace std; map<string, string> mp; void Login(string account, string passwd); void Create(string account, string passwd); int main() { int n; char c; string account, passwd; cin >> n; for (int i = 0 ; i < n; i++) { getchar(); cin >> c >> account >> passwd; switch (c) { case 'L': Login(account, passwd); break; case 'N': Create(account, passwd); break; } } return 0; } void Login(string account, string passwd) { if (mp.find(account) != mp.end()) { if (passwd == mp[account]) { cout << "Login: OK" << endl; } else { cout << "ERROR: Wrong PW" << endl; } } else { cout << "ERROR: Not Exist" << endl; } } void Create(string account, string passwd) { if (mp.find(account) != mp.end()) { cout << "ERROR: Exist" << endl; } else { mp[account] = passwd; cout << "New: OK" << endl; } }
map如果在栈中定义,地址作为函数参数传到函数中,函数中用map型指针接收的思路,还没有通过。
用创建链表的方式代替map功能
#include<cstdio> #include<cstdlib> #include<cstring> const int maxn = 100010; typedef struct ListNode* Position; typedef struct Table* HashTable; struct ListNode { char account[15]; char password[20]; Position next; }; struct Table { Position List; int TableSize; }; HashTable createTable(int size); int NextPrime(int size); void Insert(HashTable H, char *account, char *password); Position Find(HashTable H, char *key); int Find_account(HashTable H, char *account); int check_password(Position P, char *password); void DestoryTable(HashTable H); int main() { int n; scanf("%d", &n); HashTable H = createTable(n * 2); char c; char account[15]; char password[20]; for (int i = 0; i < n; i++) { getchar(); scanf("%c %s %s", &c, account, password); switch(c) { case 'N': if (-1 == Find_account(H, account)) { Insert(H, account, password); printf("New: OK\n"); } else { printf("ERROR: Exist\n"); } break; case 'L': if (1 == Find_account(H, account)) { if (1 == check_password(Find(H, account), password)) { printf("Login: OK\n"); } else { printf("ERROR: Wrong PW\n"); } } else { printf("ERROR: Not Exist\n"); } break; } } //DestoryTable(H); return 0; } HashTable createTable(int size) { //在堆区申请一个Table大小的内存,转化成该结构体指针类型返回 HashTable H = (HashTable)malloc(sizeof (struct Table)); H->TableSize = NextPrime(size); H->List = (Position)malloc(sizeof(struct ListNode) * H->TableSize); for (int i = 0 ; i < H->TableSize; i++) { H->List[i].next = NULL; } return H; } int NextPrime(int size) { for (int tmpSize = size; ; tmpSize++) { int i = 2; for ( ; i * i <= tmpSize; i++) { if (tmpSize % i == 0) { break; } } if (i * i > tmpSize) { return tmpSize; } } } void Insert(HashTable H, char *account, char *password) { int h = atoi(account+3) % H->TableSize; /*Position P = H->List[h].next; while(P && strcmp(P->account, account)) { P = P->next; }*/ Position P = Find(H, account); if (P == NULL) //头插法 { Position tmp = (Position)malloc(sizeof (struct ListNode)); strcpy(tmp->account, account); strcpy(tmp->password, password); tmp->next = H->List[h].next; H->List[h].next = tmp; } } Position Find(HashTable H, char *key) { int h = atoi(key+3) % H->TableSize; Position P = H->List[h].next; while( P != NULL && strcmp(P->account, key)) { //节点存在,并且节点的账户不是传进来的key P = P->next; } return P; } int Find_account(HashTable H, char *account) { Position P = Find(H, account); if (P) { return 1; } else { return -1; } } int check_password(Position P, char *password) { if (0 == strcmp(P->password, password)) { return 1; } else { return 0; } } void DestoryTable(HashTable H) { for (int i = 0; i < H->TableSize; i++) { Position P = H->List[i].next; while (P) { Position tmp = P->next; free(P); P = tmp; } } free(H->List); free(H); }