PAT(甲级)1078
2015-09-29 10:44
323 查看
1078. Hashing (25)
时间限制100 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
The task of this problem is simple: insert a sequence of distinct positive integers into a hash table, and output the positions of the input numbers. The hash function is defined to be "H(key) = key % TSize" where TSize is the maximum size of the hash table.
Quadratic probing (with positive increments only) is used to solve the collisions.
Note that the table size is better to be prime. If the maximum size given by the user is not prime, you must re-define the table size to be the smallest prime number which is larger than the size given by the user.
Input Specification:
Each input file contains one test case. For each case, the first line contains two positive numbers: MSize (<=104) and N (<=MSize) which are the user-defined table size and the number of input numbers, respectively. Then N distinct
positive integers are given in the next line. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print the corresponding positions (index starts from 0) of the input numbers in one line. All the numbers in a line are separated by a space, and there must be no extra space at the end of the line. In case it is impossible to insert the
number, print "-" instead.
Sample Input:
4 4 10 6 4 15
Sample Output:
0 1 4 -
#include <cstdio> #include <cmath> #include <ctime> #define SIZE 10004 #define MAX 2000 using namespace std; bool hash[SIZE]; int prime[MAX]; bool isprime(int &n,int prime[],int &k){ int i =0; for(i=0;i<k;i++) if(n%prime[i] == 0) return false; return true; } int makeprime(int n,int prime[],int &k) { k=0; int i; prime[k++] =2; int limit=sqrt(n); for(i=3;i<=limit;i+=2) if(isprime(i,prime,k)) prime[k++] = i; int k1=k; for (i; i <= n; i += 2){ if (isprime (i, prime, k1)) prime[k++] = i; } if(prime[k-1]<n){ for(i;;i+=2) if(isprime(i,prime,k)) {prime[k++] = i; break;} } return prime[k-1]; } int main() { int msize, n, i; int k; scanf ("%d%d", &msize, &n); msize = makeprime (msize, prime, k); for(i=0;i<n;i++) hash[i]=false; scanf("%d",&k); hash[k%msize]=true; printf("%d",k%msize); for (i = 0; i < n-1; i++) { scanf("%d",&k); int index0 = k%msize; int j; for(j=0;j<n;j++){ int index1 =(index0+j*j)%msize; if(!hash[index1]){ hash[index1] = true; printf(" %d",index1); break; } } if(j==n) printf(" -"); } printf("\n"); return 0; }
相关文章推荐
- vsftp与mysql结合进行用户验证
- sql注入的问题,PreparedStatement
- 【TCO 2012】2A EvenPaths
- android学习各种bug(5)
- C语言 工具栏创建
- 工厂模式实现仓储模版
- 求最大公约数和最小公倍数
- iOS UI进阶-2.0 CALayer
- 基本数据结构(2)——算法导论(12)
- 素数表
- PAT(甲级)1077
- numpy argsort排序函数
- tar命令详解
- JDBC数据库的连接基本知识点
- 接口测试
- 2.强制类型转换
- iOS 调试Log
- MakeFile
- dispatch_async 与dispatch_sync的区别
- HttpWebRequest用法