Codeforces Round #609 (Div. 2) B.Modulo Equality（思维）

time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a positive integer m and two integer sequence: a=[a1,a2,…,an] and b=[b1,b2,…,bn]. Both of these sequence have a length n.

Permutation is a sequence of n different positive integers from 1 to n. For example, these sequences are permutations: [1], [1,2], [2,1], [6,7,3,4,1,2,5]. These are not: [0], [1,1], [2,3].

You need to find the non-negative integer x, and increase all elements of ai by x, modulo m (i.e. you want to change ai to (ai+x)modm), so it would be possible to rearrange elements of a to make it equal b, among them you need to find the smallest possible x.

In other words, you need to find the smallest non-negative integer x, for which it is possible to find some permutation p=[p1,p2,…,pn], such that for all 1≤i≤n, (ai+x)modm=bpi, where ymodm — remainder of division of y by m.

For example, if m=3, a=[0,0,2,1],b=[2,0,1,1], you can choose x=1, and a will be equal to [1,1,0,2] and you can rearrange it to make it equal [2,0,1,1], which is equal to b.

Input
The first line contains two integers n,m (1≤n≤2000,1≤m≤109): number of elemens in arrays and m.

The second line contains n integers a1,a2,…,an (0≤ai<m).

The third line contains n integers b1,b2,…,bn (0≤bi<m).

It is guaranteed that there exists some non-negative integer x, such that it would be possible to find some permutation p1,p2,…,pn such that (ai+x)modm=bpi.

Output
Print one integer, the smallest non-negative integer x, such that it would be possible to find some permutation p1,p2,…,pn such that (ai+x)modm=bpi for all 1≤i≤n.

Examples
inputCopy
4 3
0 0 2 1
2 0 1 1
outputCopy
1
inputCopy
3 2
0 0 0
1 1 1
outputCopy
1
inputCopy
5 10
0 0 0 1 2
2 1 0 0 0
outputCopy
0
两种思维：
对a序列b序列对x进行暴力匹配，直至找到满足条件的x；
对b序列进行匹配，在O(n)内找到唯一的x

显然选择第二种解决办法

#include <bits/stdc++.h>
using namespace std;
typedef long long ll; 
ll a[2050],b[2050],c[2050];
void check()
{
	ll n,m,i,j,x,minx;
	scanf("%lld%lld",&n,&m);
	for(i=0;i<n;i++)
		scanf("%lld",&a[i]);
	for(i=0;i<n;i++)
		scanf("%lld",&b[i]);
	sort(b,b+n);
	minx=m;
	for(i=0;i<n;i++)
	{
		x=(a[0]-b[i]+m)%m;
		for(j=0;j<n;j++)
			c[j]=(a[j]+x)%m;
		sort(c,c+n);
		for(j=0;j<n;j++)
		{
			if(c[j]!=b[j])
				break;
			if(j==n-1)
					minx=min(x,minx);
		}
	}
	printf("%lld",minx);
	return;
}
int main()
{
    check();
    return 0;
}