Codeforces Round #383 (Div. 2) A,B,C,D 循环节，标记，暴力，并查集+分组背包

A. Arpa’s hard exam and Mehrdad’s naive cheat

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.

Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given n, print the last digit of 1378n.

Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.

Input

The single line of input contains one integer n (0  ≤  n  ≤  109).

Output

Print single integer — the last digit of 1378n.

Examples

input

1

output

8

input

2

output

4

Note

In the first example, last digit of 13781 = 1378 is 8.

In the second example, last digit of 13782 = 1378·1378 = 1898884 is 4.

题意：给你一个n，求 1387^n%10;(^为次方）

思路：循环节，特判0，或者快速幂；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=1e5+,M=1e6+,inf=1e9+;
const ll INF=1e18+,mod=;
int quick(int a,int b,int mod)
{
    int ans=;
    while(b)
    {
        if(b&)ans*=a,ans%=mod;
        b>>=;
        a=(a*a)%mod;
    }
    return ans;
}
int main()
{
    int n;
    scanf("%d",&n);
    printf("%d\n",quick(,n,));
    return ;
}

B. Arpa’s obvious problem and Mehrdad’s terrible solution

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are some beautiful girls in Arpa’s land as mentioned before.

Once Arpa came up with an obvious problem:

Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n) such that , where  is bitwise xoroperation (see notes for explanation).

Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.

Input

First line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the array and the integer x.

Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array.

Output

Print a single integer: the answer to the problem.

Examples

input

2 3
1 2

output

1

input

6 1
5 1 2 3 4 1

output

2

Note

In the first sample there is only one pair of i = 1 and j = 2.  so the answer is 1.

In the second sample the only two pairs are i = 3, j = 4 (since ) and i = 1, j = 5 (since ).

A bitwise xor takes two bit integers of equal length and performs the logical xor operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. You can read more about bitwise xor operation here: https://en.wikipedia.org/wiki/Bitwise_operation#XOR.

题意：给你n个数，求ai^aj==x（1<=i<j<=n,^为异或符号)，求i，j的总对数；

思路：标记，将式子转换一下，ai==x^aj；枚举a[i],标记a[j]得到答案；

　　   坑点：1：当x为0的时候，ai==aj，所以对答案的贡献是flag[ai]*(flag[ai]-1);

　　　　　　2：处理过程中，答案会超过int；

　　　　　　3：两个数的异或可能超过1e5；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=1e6+,M=4e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
ll a[N],flag[N];
int main()
{
    int n,x;
    scanf("%d%d",&n,&x);
    for(int i=;i<=n;i++)
        scanf("%lld",&a[i]),flag[a[i]]++;
    ll ans=;
    for(int i=;i<=;i++)
    {
        int z=i^x;
        if(z==i)
            ans+=(flag[z]*(flag[z]-));
        else
            ans+=flag[z]*flag[i];
    }
    printf("%lld\n",ans/);
    return ;
}

C. Arpa's loud Owf and Mehrdad's evil plan

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

As you have noticed, there are lovely girls in Arpa’s land.

People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.

Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.

The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeatedt times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called theJoon-Joon of the round. There can't be two rounds at the same time.

Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.

Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).

Input

The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.

The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.

Output

If there is no t satisfying the condition, print -1. Otherwise print such smallest t.

Examples

input

4
2 3 1 4

output

3

input

4
4 4 4 4

output

-1

input

4
2 1 4 

题意：找到一个最小的t，x走t步走到y，y也必须走t步走到x，x->a[x]->a[a[x]]->....->y;

思路：当a[i]为1-n且每个数都不相等,否则为-1；

　　　暴力找环，当环的大小x为奇数时，这个环的贡献为x；

　　　　　　　　当环的大小x为偶数时，这个环的贡献为x/2;

　　　求所有环的贡献的lcm；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define esp 0.00000000001
const int N=1e3+,M=1e6+,inf=1e9;
const ll INF=1e18+;
int a[N],flag[N];
ll ans=;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=;i<=n;i++)
    scanf("%d",&a[i]);
    for(int i=;i<=n;i++)
    {
        if(flag[a[i]])return puts("-1");
        flag[a[i]]=;
    }
    memset(flag,,sizeof(flag));
    for(int i=;i<=n;i++)
    {
        if(flag[i])continue;
        ll hh=;
        int x=i;
        while(!flag[x])
        {
            flag[x]=;
            hh++;
            x=a[x];
        }
        if(hh&)
            ans=ans*hh/__gcd(hh,ans);
        else
        ans=ans*(hh/)/__gcd(ans,hh/);
    }
    printf("%lld\n",ans);
    return ;
}

D. Arpa's weak amphitheater and Mehrdad's valuable Hoses

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Just to remind, girls in Arpa's land are really nice.

Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.

Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.

Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.

Input

The first line contains integers n, m and w (1  ≤  n  ≤  1000, , 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.

The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.

The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.

The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xiand yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.

Output

Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.

Examples

input

3 1 5
3 2 5
2 4 2
1 2

output

6

input

4 2 11
2 4 6 6
6 4 2 1
1 2
2 3

output

7

Note

In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.

In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.

题意：n个人去参加派对，m对关系，w为你可以带走的重量；

　　　每个人手中有一个礼物，一个重量w，一个美丽程度b；

　　　m对关系两个人是朋友；

　　　对于多个朋友，要么带走全部礼物，要么带走一个，要么都不带走；

思路：并查集+分组背包；

　　　并查集分组；

　　 对于分组背包，每组中再加入一个全部礼物；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define esp 0.00000000001
const int N=1e3+,M=1e6+,inf=1e9;
const ll INF=1e18+;
int fa[N];
int Find(int x)
{
    return x==fa[x]?x:fa[x]=Find(fa[x]);
}
void update(int x,int y)
{
    int u=Find(x);
    int v=Find(y);
    if(u!=v)
    {
        fa[u]=v;
    }
}
vector<pair<int,int> >f[N];
ll sumw[N],sumb[N];
int si;
int w[N],b[N],flag[N];
ll dp[N][N];
int main()
{
    int n,val,m;
    scanf("%d%d%d",&n,&m,&val);
    for(int i=;i<=n;i++)
        fa[i]=i;
    for(int i=;i<=n;i++)
        scanf("%d",&w[i]);
    for(int i=;i<=n;i++)
        scanf("%d",&b[i]);
    for(int i=;i<=m;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        update(u,v);
    }
    si=;
    for(int i=;i<=n;i++)
    {
        int yy=Find(i);
        if(!flag[yy])
        {
            si++;
            flag[yy]=si;
            f[si].push_back(make_pair(w[i],b[i]));
            sumw[si]+=w[i];
            sumb[si]+=b[i];
        }
        else
        {
            f[flag[yy]].push_back(make_pair(w[i],b[i]));
            sumw[flag[yy]]+=w[i];
            sumb[flag[yy]]+=b[i];
        }
    }
    for(int i=;i<=n;i++)
    {
        if(flag[i])
        {
            f[flag[i]].push_back(make_pair(sumw[flag[i]],sumb[flag[i]]));
        }
    }
    /*for(int i=1;i<=si;i++)
    {
        for(int j=0;j<f[i].size();j++)
        cout<<f[i][j].first<<" "<<f[i][j].second<<" "<<endl;
        cout<<"~~~~~~~~~~~~"<<endl;
    }*/
    for(int i=;i<=si;i++)
    {
        for(int j=;j<f[i].size();j++)
        {
            for(int k=val;k>=;k--)
            {
                dp[i][k]=max(dp[i-][k],dp[i][k]);
                if(k>=f[i][j].first)
                dp[i][k]=max(dp[i][k],dp[i-][k-f[i][j].first]+f[i][j].second);
            }
        }
    }
    printf("%lld\n",dp[si][val]);
    return ;
}