POJ 2769 Reduced ID Numbers 同余定理（暴力）

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本文探讨了大学教师T.Chur如何通过找到最小的正整数m来解决学生身份号码模数问题，确保同一组内所有身份号码模m后的结果各不相同。

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Reduced ID Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8596		Accepted: 3448

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 = 10 ⁶-1. 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

Sample Output

1
8

Source

Northwestern Europe 2005

每个学生都有一个SIN，但是范围太大，求在每个组里能找到最小的正整数m，使得当前组内的所有数对模m均布同余。

枚举每一个数，暴力搜。

//844K	454MS	
#include<stdio.h>
#include<string.h>
int s[100007];
bool vis[100007];
int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int m,i;
        memset(s,0,sizeof(s));
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
            scanf("%d",&s[i]);
        for(i=1;;i++)
        {
            int flag=0;
            memset(vis,0,sizeof(vis));
            for(int j=1;j<=m;j++)
            {
                if(vis[s[j]%i])//对i取余得到的结果在次数之前已经有过
                {
                    flag=1;//标记此i不是所求，break
                    break;
                }
                vis[s[j]%i]=1;//否则标记访问过
            }
            if(!flag)break;
        }
        printf("%d\n",i);
    }
    return 0;
}