POJ 2769 reduced ID Numbers (same remainder)

Reduced ID Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 9153		Accepted: 3675

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 this within the group all SINs reduced modulo m ar E unique.

Input

On the first line of the input was a single positive integer N, telling the number of the 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 is distinct, though not necessarily sorted.

Output

For each test case, the output one line containing the smallest modulus m, such this all SINs reduced modulo m is distinct.

Sample Input

211248663124866111111987651

Sample Output

18

the main topic: to the G students of the school number, the smallest m, so that each student's school number on M is not equal. Obviously, the problem must have a solution, the range [G,max_sin], so the direct enumeration on the line, and then use the is[] array to weigh.

#include <stdio.h> #include <string.h>const int maxn=1000000+10;bool is[maxn];int a[330];int main () {int t,n , i,m;scanf ("%d", &t), while (t--) {scanf ("%d", &n), for (i=0;i<n;i++) scanf ("%d", &a[i]); for (m=n;; m++) {for (i=0;i<m;i++) is[i]=0;for (i=0;i<n;i++) {if (!is[a[i]%m]) Is[a[i]%m]=1;else break;} if (i==n) break;} printf ("%d\n", m);} return 0;}

POJ 2769 reduced ID Numbers (same remainder)