CF1500B. Two chandeliers（扩展欧几里得算法）

https://codeforces.com/contest/1500/problem/B

B. Two chandeliers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya is a CEO of a big construction company. And as any other big boss he has a spacious, richly furnished office with two crystal chandeliers. To stay motivated Vasya needs the color of light at his office to change every day. That's why he ordered both chandeliers that can change its color cyclically. For example: red – brown – yellow – red – brown – yellow and so on.

There are many chandeliers that differs in color set or order of colors. And the person responsible for the light made a critical mistake — they bought two different chandeliers.

Since chandeliers are different, some days they will have the same color, but some days — different. Of course, it looks poor and only annoys Vasya. As a result, at the kk-th time when chandeliers will light with different colors, Vasya will become very angry and, most probably, will fire the person who bought chandeliers.

Your task is to calculate the day, when it happens (counting from the day chandeliers were installed). You can think that Vasya works every day without weekends and days off.

Input

The first line contains three integers nn, mm and kk (1≤n,m≤5000001≤n,m≤500000; 1≤k≤10121≤k≤1012) — the number of colors in the first and the second chandeliers and how many times colors should differ to anger Vasya.

The second line contains nn different integers aiai (1≤ai≤2⋅max(n,m)1≤ai≤2⋅max(n,m)) that describe the first chandelier's sequence of colors.

The third line contains mm different integers bjbj (1≤bi≤2⋅max(n,m)1≤bi≤2⋅max(n,m)) that describe the second chandelier's sequence of colors.

At the ii-th day, the first chandelier has a color axax, where x=((i−1)modn)+1)x=((i−1)modn)+1) and the second one has a color byby, where y=((i−1)modm)+1)y=((i−1)modm)+1).

It's guaranteed that sequence aa differs from sequence bb, so there are will be days when colors of chandeliers differs.

Output

Print the single integer — the index of day when Vasya will become angry.

Examples

input

Copy

4 2 4
4 2 3 1
2 1

output

Copy

5

input

Copy

3 8 41
1 3 2
1 6 4 3 5 7 2 8

output

Copy

47

input

Copy

1 2 31
1
1 2

output

Copy

62

Note

In the first example, the chandeliers will have different colors at days 11, 22, 33 and 55. That's why the answer is 55.

参考：

https://blog.youkuaiyun.com/ojzha/article/details/79269727

https://blog.youkuaiyun.com/destiny1507/article/details/81751168

扩展欧几里得算法主要用于求类似ax+by=gcd(a,b)方程的一组（x,y）的整数解，类似辗转相除法，只不过这里用类似辗转相除法把方程求解变换成另一个系数更小的方程，a*x+b*y==gcd(a,b)==b*x'+(a%b)*y'

最后求得特解，可处理之后变成最小非负整数解，先将a，b互质处理a=a/gcd，b=b/gcd，求得解后再x=(x*v%b+b)%b,(因为a*x要以a*b尺度增减，才能有整数y解);

此题先求出两列数组长度n，m的最小公倍数长度s时，有多少不同的数c。可知总序列前面有（k-1）/c个s长度循环。用二分查出剩余需要多少长度，需要预先用扩展欧几里得算出每个数在哪个位置相遇。注意要处理扩展欧几里得最小满足题目情况的整数解。

import java.util.*;
import java.io.*;

public class Main {
    public static void main(String args[]) {new Main().run();}

    FastReader in = new FastReader();
    PrintWriter out = new PrintWriter(System.out);
    void run(){
        work();
        out.flush();
    }
    long mod=998244353;
    long gcd(long a,long b) {
        return a==0?b:gcd(b%a,a);
    }
    /**
     * a*x+b*y==gcd(a,b)
     * 变换为求解方程b*x'+(a%b)*y' => a*y'+b*(x'-[a/b]*y') => x=y',y=x'-[a/b]*y'
     */
    void exgcd(long a,long b,long[] A) {//f代表正负号
        if(b==0){
            A[0]=1;
            A[1]=0;
            return;
        }
        exgcd(b,a%b,A);
        long x=A[1];
        long y=A[0]-a/b*A[1];
        A[0]=x;
        A[1]=y;
    }

    /**
     * 求a*x+b*y=v的解，v必定是gcd(a,b)的倍数，返回最小非负整数x
     * @param a a>0
     * @param b b>0
     * @param v v>0
     * @return 返回最小非负整数x
     */
    long[] exgcdArr=new long[2];
    long exgcdVal(long a,long b ,long v){
        if(v==0){
            return 0;
        }
        long gcd=gcd(a,b);
        if(v%gcd!=0){
            return -1;
        }
        //a,b互质处理
        a=a/gcd;
        b=b/gcd;
        v=v/gcd;
        exgcd(a,b,exgcdArr);
        //a*x 以最小公倍数尺度(a*c)增减
        exgcdArr[0]=((exgcdArr[0]%b)*v%b+b)%b;//找到最小的非负整数系数
        return exgcdArr[0];
    }
    int n,m;
    long gcd;
    long[] rec;
    void work() {
        n=ni();m=ni();
        long k=nl();
        int[] A=nia(n);
        int[] B=nia(m);
        int[] rec1=new int[1000002];
        int[] rec2=new int[1000002];
        rec=new long[1000002];//记录数组A的数对应数组B的数相遇时的距离
        Arrays.fill(rec1,-1);
        Arrays.fill(rec2,-1);
        for(int i=0;i<n;i++){
            rec1[A[i]]=i+1;
        }
        for(int i=0;i<m;i++){
            rec2[B[i]]=i+1;
        }
        gcd=gcd(n,m);
        long s=(long)n*m/gcd;
        long ret=0;
        long c=0;
        for(int i=0;i<1000002;i++){
            if(rec1[i]>=0){
                int idx1=rec1[i];
                int idx2=rec2[i];
                if(idx2==-1)continue;
                if(Math.abs(idx2-idx1)%gcd==0){
                    if(idx1>idx2){
                        long v=idx1-idx2;
                        long a=m,b=n;
                        long x=exgcdVal(a,b,v);
                        long s1=a*x;
                        //s1>v的最小s1值,以最小公倍数相加
                        long v2=a/gcd(a,b)*b;
                        if(v>s1){
                            s1+=((v-s1-1)/(v2)+1)*v2;
                        }
                        rec[i]=s1+idx2;
                    }else{
                        long v=idx2-idx1;
                        long a=n,b=m;
                        long x=exgcdVal(a,b,v);
                        long s1=a*x;
                        long v2=a/gcd(a,b)*b;
                        if(v>s1){
                            s1+=((v-s1-1)/(v2)+1)*v2;
                        }
                        rec[i]=s1+idx1;
                    }
                    c++;
                }
            }
        }
        long c1=s-c;
        ret+=(k-1)/c1*s;
        k%=c1;
        if(k==0){
            k=c1;
        }
        long l=1,r=s;
        while(l<r){
            long mid=(l+r)/2;
            if(count(mid)<k){
                l=mid+1;
            }else{
                r=mid;
            }
        }
//        if(n==483413){
//            out.println(k+":"+c1+":"+s+":"+gcd+","+ret+";;"+l);
//        }
        out.println(ret+l);
    }

    private long count(long d) {
        long c=0;
        for(int i=0;i<1000001;i++){
            if(rec[i]>0&&rec[i]<=d){
                c++;
            }
        }
        return d-c;
    }

    //input
    @SuppressWarnings("unused")
    private ArrayList<Integer>[] ng(int n, int m) {
        ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
        for(int i=0;i<n;i++) {
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++) {
            int s=in.nextInt()-1,e=in.nextInt()-1;
            graph[s].add(e);
            graph[e].add(s);
        }
        return graph;
    }

    private ArrayList<long[]>[] ngw(int n, int m) {
        ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
        for(int i=0;i<n;i++) {
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++) {
            long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
            graph[(int)s].add(new long[] {e,w});
            graph[(int)e].add(new long[] {s,w});
        }
        return graph;
    }

    private int ni() {
        return in.nextInt();
    }

    private long nl() {
        return in.nextLong();
    }

    private String ns() {
        return in.next();
    }

    private long[] na(int n) {
        long[] A=new long[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextLong();
        }
        return A;
    }
    private int[] nia(int n) {
        int[] A=new int[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextInt();
        }
        return A;
    }
}

class FastReader
{
    BufferedReader br;
    StringTokenizer st;

    public FastReader()
    {
        br=new BufferedReader(new InputStreamReader(System.in));
    }


    public String next()
    {
        while(st==null || !st.hasMoreElements())//回车，空行情况
        {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    public int nextInt()
    {
        return Integer.parseInt(next());
    }

    public long nextLong()
    {
        return Long.parseLong(next());
    }
}