poj 2769 Reduced ID Numbers

最新推荐文章于 2018-08-08 11:16:57 发布

转载最新推荐文章于 2018-08-08 11:16:57 发布 · 166 阅读

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原文链接：http://www.cnblogs.com/w0w0/archive/2011/11/22/2258653.html

本文介绍了一个算法问题——ReducedIDNumbers，该问题旨在找到一组学生ID号码中能使所有ID变得唯一的最小正整数模数。通过输入一组学生ID，程序会输出使得这些ID在取模后仍能保持唯一性的最小模数。

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Reduced ID Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7346		Accepted: 2971

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

#include<iostream>
#include<algorithm>
using namespace std;

int comp(int &a,int &b)
{
return a<b;
}

int main()
{
int nums[300];
int modu[300];
int g;
int n;
int divi;
int i;
    cin>>n;
while(n--)
    {
        cin>>g;
for(i=0;i<g;i++)
            cin>>nums[i];
for(divi=g;;divi++)
        {
for(i=0;i<g;i++)
            {
                modu[i]=nums[i]%divi;
            }
            sort(modu,modu+g,comp);
for(i=0;i<g-1;i++)
            {
if(modu[i]==modu[i+1])
break;
            }
if(i==g-1)
            {
                cout<<divi<<endl;
break;
            }
        }
    }
return 0;
}

转载于:https://www.cnblogs.com/w0w0/archive/2011/11/22/2258653.html