Sicily 1190. Reduced ID Numbers 简单哈希

最新推荐文章于 2019-12-02 22:08:05 发布

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本文探讨了大学讲师T.Chur在管理不同组的学生时遇到的问题，即如何找到最小的正整数m，使得每个学生群体中所有学生ID在模m运算后仍保持唯一性。通过输入描述和输出需求，读者可以了解如何解决此类问题并优化学生分组策略。

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Time Limit: 2 secs, Memory Limit: 32 MB

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

︿(￣︶￣)︿ Just do it!

#include <iostream>
#include <cstring>

using namespace std;

int main()
{
    int T, N, index, m, SIN[310];
    bool HASH[99999];
    cin >> T;
    while (T--)
    {
        cin >> N;
        for (int i = 1; i <= N; i++) {
            cin >> SIN[i];
        }
        m = N;  // 起始值为N  \\Think about it！
        index = 1;
        while (index <= N)
        {   
            // 不可设置为index < N  \\Think about it！
            memset(HASH, false, sizeof(HASH));
            for (index = 1; index <= N; index++)
                if (!HASH[SIN[index] % m])
                    // 标记该位置已被占用
                    HASH[SIN[index] % m] = true;
                else
                    // 证明有两个元素模值相同 => Fail
                    break;      
            m++;
        }
        cout << --m << endl;
    }
    return  0;
}