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  • POJ 2769 Reduced ID Numbers

    思路: 枚举

    Reduced ID Numbers
    Time Limit: 2000MS Memory Limit: 65536K
    Total Submissions: 8847 Accepted: 3552
    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
    Source

    Northwestern Europe 2005

    <span style="color:#3333ff;">/*********************************************
            author    : Grant Yuan
            time      : 2014/8/21 14:45
            algorithm : Basic Math
            source    : POJ 2769
    **********************************************/
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #define MAX 100000
    
    using namespace std;
    bool flag[MAX];
    int a[307];
    int b[307];
    
    void init(int n)
    {
        for(int i=0;i<n;i++)
        {
            flag[b[i]]=false;
            b[i]=0;
        }
    }
    
    int main()
    {
        int t,ans;
        scanf("%d",&t);
        while(t--){
            int n;
            scanf("%d",&n);
            memset(flag,0,sizeof(flag));
            memset(b,0,sizeof(b));
            for(int i=0;i<n;i++)
            {
                scanf("%d",&a[i]);
            }
            for(int i=n;;i++)
            {
                init(n);
                bool f=true;
               for(int j=0;j<n;j++)
               {
                  int m=a[j]%i;
                  if(flag[m]) {f=false;break;}
                  else {flag[m]=true; b[j]=m;}
               }
               if(f) {ans=i;break;}
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
    </span>


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  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254446.html
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