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

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

做法： $n^2$暴力枚举，对于a中第一个元素，肯定会映射到b中去，假设映射到 $b_i(1\leq i \leq n)$，计算出差值（可能会有两个），判断是否合法，并不断对答案取min
代码：

/**
 *  Author1: low-equipped w_udixixi
 *  Author2: Sher丶lock
 *  Date :2019-12-21
 **/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<queue>
#include<map>
#define ll long long
#define pb push_back
#define rep(x,a,b) for (int x=a;x<=b;x++)
#define repp(x,a,b) for (int x=a;x<b;x++)
#define W(x) printf("%d\n",x)
#define WW(x) printf("%lld\n",x)
#define pi 3.14159265358979323846
#define mem(a,x) memset(a,x,sizeof a)
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
using namespace std;
const int maxn=2e6+7;
const int INF=1e9;
const ll INFF=1e18;
int a[maxn],b[maxn],n,m,ans=INF;
int main()
{
    scanf("%d%d",&n,&m);
    rep(i,1,n)scanf("%d",&a[i]);
    rep(i,1,n){scanf("%d",&b[i]);b[i]+=m;}
    sort(a+1,a+1+n);sort(b+1,b+1+n);
    rep(i,1,n)
    {
        int x=b[i]-a[1];
        int u=1,d=i;
        bool mark=true;
        while(u<=n)
        {
            if (b[d]-a[u]!=x&&b[d]-a[u]!=x-m)
            {
                mark=false;
                break;
            }
            u++;d=(d+1)%(n+1);
            if (d==0)d++;
        }
        if (x<m&&mark)ans=min(ans,x);
        else if (mark)ans=min(ans,x-m);
    }
    W(ans);
}

sb题，恶心了我一个下午
我呸