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2013
11-12

POJ 2769 Reduced ID Numbers [解题报告] Java

Reduced ID Numbers

问题描述 :

T. Chur teaches various groups of students at university U. Every U-student has a unique Student Identification Number (SIN). A SIN s is an integer in the range 0 ≤ s ≤ MaxSIN with MaxSIN = 106-1. T. Chur finds this range of SINs too large for identification within her groups. For each group, she wants to find the smallest positive integer m, such that within the group all SINs reduced modulo m are unique.

输入:

On the first line of the input is a single positive integer N, telling the number of test cases (groups) to follow. Each case starts with one line containing the integer G (1 ≤ G ≤ 300): the number of students in the group. The following G lines each contain one SIN. The SINs within a group are distinct, though not necessarily sorted.

输出:

For each test case, output one line containing the smallest modulus m, such that all SINs reduced modulo m are distinct.

样例输入:

2
1
124866
3
124866
111111
987651

样例输出:

1
8

解题代码:

//* @author  [email protected]
import java.util.Hashtable;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Hashtable< Integer,Integer> table;
        Scanner scan=new Scanner(System.in);
        int M[]=new int[300];
        int n=scan.nextInt(),k=0,m=1;
        boolean band=false;
        for(int i=0;i< n;i++){
            k=scan.nextInt();
            band=true;
            table=new Hashtable< Integer, Integer>();
            for(int j=0;j< k;j++){
                M[j]=scan.nextInt();
            }
            m=k;
            while(true){
                band=true;
                for(int j=0;j< k;j++){
                    if(table.get((M[j]%m))==null){
                        table.put((M[j]%m), 0);
                    }
                    else {
                        band=false;
                        break;
                    }
                }
                if(band==true) break;
                table.clear();
                m++;
            }
            System.out.println(m);
        }
    }
}