算法学习【14】—— 1190. Reduced ID Numbers

寻找最小模数

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本文介绍了一个算法问题，即如何找到一组学生ID号码中使得所有ID模m后仍保持唯一的最小正整数m。通过暴力搜索的方法，实现了这一目标并给出了具体的C语言实现代码。

题目来源：http://soj.me/1190

1190. Reduced ID Numbers

Constraints

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

Sample Output

1
8

思路：

这个看懂题目应该就OK的了，太久没敲代码，结果WA了一次student数组没有剩余。暴力搜索吧。

代码：

/*
Problem link: http://soj.me/1190
Author: BetaBin
Date: 2012/07/25
*/
#include <stdio.h>
#include <memory.h>

int students[302];
bool buckets[1000000];
int studentNum;
int groupNum;
int answer;

bool Test(int target)
{
    int remainder;
    memset(buckets, false, sizeof(bool) * answer);
    for(int i = 0; i < studentNum; ++i)
    {
        remainder = students[i] % target;
        if(buckets[remainder])
        {
            return false;
        }
        else
        {
            buckets[remainder] = true;
        }
    }
    return true;
}

int main()
{
    scanf("%d", &groupNum);
    while(groupNum--)
    {
        scanf("%d", &studentNum);
        for(answer = 0; answer < studentNum; ++answer)
        {
            scanf("%d", &students[answer]);
        }
        while(!Test(answer))
        {
            ++answer;
        }
        printf("%d\n", answer);
    }
    return 0;
}