POJ2769 Reduced ID Numbers[同余]

最新推荐文章于 2016-07-24 21:00:00 发布

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本文介绍了一种算法问题，即如何找到一个最小的正整数m，使得一组学生标识号（SIN）对m取模后的结果都是唯一的。通过遍历每个可能的m值，并使用标记数组来检查是否有重复的余数，最终确定满足条件的最小m。

Reduced ID Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 9813		Accepted: 3923

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

题意:

找到与G个数取余的结果都不相同的数

从1开始循环开始,因为一定是有结果的,所以循环可以不设置停止条件,找到结果之后,终止循环即可

每次更新mark数组的时候,只需要吧0到i-1的值归零,因为余数只会在0到i-1的范围内.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
using namespace std;
const int N=310;
const int M=1e6;
int num[N];
bool mark[M];
int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        int n;
        scanf("%d",&n);
        for (int i=0 ; i<n ; i++)
            scanf("%d",&num[i]);
        int ans=1;
        for (int i=1 ; ; i++)
        {
            for (int j=0 ; j<i ; j++)//余数只会取到比i小的数
                mark[j]=0;
            bool flag=0;
            for (int j=0 ; j<n ; j++)
            {
                if (!mark[num[j]%i])
                    mark[num[j]%i]=1;
                else                     
                {
                    flag=1;
                    break;
                }
            }
            if (!flag)
            {
                ans=i;
                break;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}