POJ 2769 Reduced ID Numbers

Reduced ID Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7494		Accepted: 3029

Description

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.

Input

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.

Output

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

Sample Input

2
1
124866
3
124866
111111
987651

Sample Output

1
8

Source

Northwestern Europe 2005

 

 

 

枚举法。。。。

注意清除标志时，只要清除用过的记录，直接memset会超时的

 

#include<stdio.h>
#include<string.h>
int a[310];
bool flag[1000010];
int set[310];
void clear(int n)
{
    for(int i=0;i<n;i++)
      flag[set[i]]=false;
}    
int main()
{
    int j;
    int T;
    int n;
    int m;
    bool ff;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%d",&a[i]);
        memset(flag,false,sizeof(flag));
        m=0;
        for(j=n;;j++)
        {
            clear(m);
            m=0;
            ff=true;
            for(int i=0;i<n;i++)
            {
                int t=a[i]%j;
                if(flag[t]==true)
                {
                    ff=false;
                    break;
                }    
                else
                {
                    flag[t]=true;
                    set[m++]=t;
                }    
            }    
            if(ff) break;
        }    
        printf("%d\n",j);
        
    }    
    return 0;
}