【取余优化】——poj2769——暴力中的优化

                                       Reduced ID Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 9795		Accepted: 3917

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

题意：求最小不同余的数

思路：直接枚举除数

注意：两点优化：（不只是输入输出）
1、枚举除数 i 的时候需要从 n（数的个数）开始。（因为余数的个数是n个不同的数，所以要求i最小为n）
2、memset函数的初始化空间需要写成除数 i 个大小。

代码如下：

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;

int id[305];
bool jug[1000005];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&id[i]);

        int find=0;

        int i;
        for(i=n;;i++)
        {
            find=1;
            memset(jug,0,sizeof(bool)*i);
            for(int j=1;j<=n;j++)
            {
                if(jug[id[j]%i]==true)
                {
                    find=0;
                    break;
                }
                jug[id[j]%i]=1;
            }
            if(find==1)
                break;
        }
        printf("%d\n",i);
    }
    return 0;
}