Codeforces Round #269 (Div. 2) A~D_html/css_WEB-ITnose

这次的CF除了最后一个都比较简单,

第一次用JAVA写CF.......

Codeforces Round #269 (Div. 2)

A. MUH and Sticks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Two polar bears Menshykov and Uslada from the St.Petersburg zoo and elephant Horace from the Kiev zoo got six sticks to play with and assess the animals' creativity. Menshykov, Uslada and Horace decided to make either an elephant or a bear from those sticks. They can make an animal from sticks in the following way:

Four sticks represent the animal's legs, these sticks should have the same length.

Two remaining sticks represent the animal's head and body. The bear's head stick must be shorter than the body stick. The elephant, however, has a long trunk, so his head stick must be as long as the body stick. Note that there are no limits on the relations between the leg sticks and the head and body sticks.

Your task is to find out which animal can be made from the given stick set. The zoo keeper wants the sticks back after the game, so they must never be broken, even bears understand it.

Input

The single line contains six space-separated integers li (1?≤?li?≤?9) ? the lengths of the six sticks. It is guaranteed that the input is such that you cannot make both animals from the sticks.

Output

If you can make a bear from the given set, print string "Bear" (without the quotes). If you can make an elephant, print string "Elephant" (w?thout the quotes). If you can make neither a bear nor an elephant, print string "Alien" (without the quotes).

Sample test(s)

input

4 2 5 4 4 4

output

Bear

input

4 4 5 4 4 5

output

Elephant

input

1 2 3 4 5 6

output

Alien

Note

If you're out of creative ideas, see instructions below which show how to make a bear and an elephant in the first two samples. The stick of length 2 is in red, the sticks of length 4 are in green, the sticks of length 5 are in blue.

简单题:

/** * Created by ckboss on 14-9-27. */import java.util.*;public class CF471A {    static int[] a = new int[6];    static int[] num = new int[10];    public static void main(String[] args){        Scanner in = new Scanner(System.in);        for(int i=0;i<6;i++){            a[i]=in.nextInt();        }        int mx=-1,sg=-1;        for(int i=0;i<6;i++){            num[a[i]]++;            if(mx<num[a[i]]){                mx=num[a[i]];                sg=a[i];            }        }        if((mx<4)==true){            System.out.println("Alien");            return ;        }        num[sg]-=4;        int x=-1,y=-1;        for(int i=1;i<=9;i++){            while(num[i]>0){                if(x==-1) x=i;                else if(y==-1) y=i;                num[i]--;            }        }        if(x==y){            System.out.println("Elephant");        }        else{            System.out.println("Bear");        }    }}

B. MUH and Important Things

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

It's time polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got down to business. In total, there are n tasks for the day and each animal should do each of these tasks. For each task, they have evaluated its difficulty. Also animals decided to do the tasks in order of their difficulty. Unfortunately, some tasks can have the same difficulty, so the order in which one can perform the tasks may vary.

Menshykov, Uslada and Horace ask you to deal with this nuisance and come up with individual plans for each of them. The plan is a sequence describing the order in which an animal should do all the n tasks. Besides, each of them wants to have its own unique plan. Therefore three plans must form three different sequences. You are to find the required plans, or otherwise deliver the sad news to them by stating that it is impossible to come up with three distinct plans for the given tasks.

Input

The first line contains integer n (1?≤?n?≤?2000) ? the number of tasks. The second line contains n integers h1,?h2,?...,?hn (1?≤?hi?≤?2000), where hi is the difficulty of the i-th task. The larger number hi is, the more difficult the i-th task is.

Output

In the first line print "YES" (without the quotes), if it is possible to come up with three distinct plans of doing the tasks. Otherwise print in the first line "NO" (without the quotes). If three desired plans do exist, print in the second line n distinct integers that represent the numbers of the tasks in the order they are done according to the first plan. In the third and fourth line print two remaining plans in the same form.

If there are multiple possible answers, you can print any of them.

Sample test(s)

input

41 3 3 1

output

YES1 4 2 3 4 1 2 3 4 1 3 2 

input

52 4 1 4 8

output

NO

Note

In the first sample the difficulty of the tasks sets one limit: tasks 1 and 4 must be done before tasks 2 and 3. That gives the total of four possible sequences of doing tasks : [1, 4, 2, 3], [4, 1, 2, 3], [1, 4, 3, 2], [4, 1, 3, 2]. You can print any three of them in the answer.

In the second sample there are only two sequences of tasks that meet the conditions ? [3, 1, 2, 4, 5] and [3, 1, 4, 2, 5]. Consequently, it is impossible to make three distinct sequences of tasks.

简单题....

/** * Created by ckboss on 14-9-27. */import java.util.*;public class CF471B {    static class Dui{        int num,id;    }    static class mycmp implements Comparator<Dui>{        public int compare(Dui a,Dui b){            return a.num-b.num;        }    }    static int n;    static int[] h = new int[2200];    static int[] used = new int[2200];    static Dui[] dui = new Dui[2200];    public static void main(String[] args){        Scanner in = new Scanner(System.in);        n=in.nextInt();        int mx=-1;        for(int i=1;i<=n;i++) {            h[i]=in.nextInt();            dui[i] = new Dui();            dui[i].num=h[i]; dui[i].id=i;            used[h[i]]++;            mx=Math.max(mx,used[h[i]]);        }        Arrays.sort(dui,1,n+1,new mycmp());        if(mx>=3){            System.out.println("YES");            int sg=-1;            for(int i=1;i<=n;i++){                if(used[h[i]]>=3){                    sg=h[i];                    break;                }            }            int a=-1,b=-1,c=-1,nt=0;            for(int i=1;i<=n;i++){                if(h[i]==sg){                    if(nt==0) a=i;                    else if(nt==1) b=i;                    else if(nt==2) c=i;                    nt++;                    if(nt==3) break;                }            }            for(int i=1;i<=n;i++){                System.out.print(dui[i].id+" ");            }            System.out.println("");            for(int i=1;i<=n;i++){                int t=dui[i].id;                if(t==a) t=b;                else if(t==b) t=a;                System.out.print(t+" ");            }            System.out.println("");            for(int i=1;i<=n;i++){                int t=dui[i].id;                if(t==c) t=b;                else if(t==b) t=c;                System.out.print(t+" ");            }            System.out.println("");        }        else if(mx==2){            int x=-1,y=-1,nt=0;            int p1=-1,p2=-1;            int q1=-1,q2=-1;            for(int i=1;i<=n;i++){                if(used[h[i]]==2){                    if(nt==0)                    {                        x=h[i];                        nt++;                    }                    else if(nt==1&&h[i]!=x)                    {                        y=h[i];                        nt++;                    }                    if(nt==2) break;                }            }            if(x==-1||y==-1){                System.out.println("NO");                return ;            }            System.out.println("YES");            for(int i=1;i<=n;i++){                if(h[i]==x){                    if(p1==-1) p1=i;                    else if(p2==-1) p2=i;                }                if(h[i]==y){                    if(q1==-1) q1=i;                    else if(q2==-1) q2=i;                }            }            for(int i=1;i<=n;i++){                System.out.print(dui[i].id+" ");            }            System.out.println("");            for(int i=1;i<=n;i++){                int t=dui[i].id;                if(t==p1) t=p2;                else if(t==p2) t=p1;                System.out.print(t+" ");            }            System.out.println("");            for(int i=1;i<=n;i++){                int t=dui[i].id;                if(t==q1) t=q2;                else if(t==q2) t=q1;                System.out.print(t+" ");            }        }        else{            System.out.println("NO");            return ;        }    }}

C. MUH and House of Cards

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev decided to build a house of cards. For that they've already found a hefty deck of n playing cards. Let's describe the house they want to make:

1. The house consists of some non-zero number of floors.
2. Each floor consists of a non-zero number of rooms and the ceiling. A room is two cards that are leaned towards each other. The rooms are made in a row, each two adjoining rooms share a ceiling made by another card.
3. Each floor besides for the lowest one should contain less rooms than the floor below.

Please note that the house may end by the floor with more than one room, and in this case they also must be covered by the ceiling. Also, the number of rooms on the adjoining floors doesn't have to differ by one, the difference may be more.

While bears are practicing to put cards, Horace tries to figure out how many floors their house should consist of. The height of the house is the number of floors in it. It is possible that you can make a lot of different houses of different heights out of n cards. It seems that the elephant cannot solve this problem and he asks you to count the number of the distinct heights of the houses that they can make using exactly n cards.

Input

The single line contains integer n (1?≤?n?≤?1012) ? the number of cards.

Output

Print the number of distinct heights that the houses made of exactly n cards can have.

Sample test(s)

input

13

output

input

output

Note

In the first sample you can build only these two houses (remember, you must use all the cards):

Thus, 13 cards are enough only for two floor houses, so the answer is 1.

The six cards in the second sample are not enough to build any house.

到10^12总共只要80W种可能的层数,鉴于CF的强大机器直接暴力....

/** * Created by ckboss on 14-9-27. */import java.util.*;public class CF471C {    static long getA(long x){        return 2*(x+1)+3*(x+1)*x/2;    }    public static void main(String[] args){        Scanner in = new Scanner(System.in);        long n = in.nextLong();        int ans=0;        for(long i=0;;i++){            long t=getA(i);            if(t<=n){                long res=n-t;                if(res%3==0){                    ans++;                }            }            else break;        }        System.out.println(ans);    }}

D. MUH and Cube Walls

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got hold of lots of wooden cubes somewhere. They started making cube towers by placing the cubes one on top of the other. They defined multiple towers standing in a line as a wall. A wall can consist of towers of different heights.

Horace was the first to finish making his wall. He called his wall an elephant. The wall consists of w towers. The bears also finished making their wall but they didn't give it a name. Their wall consists of n towers. Horace looked at the bears' tower and wondered: in how many parts of the wall can he "see an elephant"? He can "see an elephant" on a segment of w contiguous towers if the heights of the towers on the segment match as a sequence the heights of the towers in Horace's wall. In order to see as many elephants as possible, Horace can raise and lower his wall. He even can lower the wall below the ground level (see the pictures to the samples for clarification).

Your task is to count the number of segments where Horace can "see an elephant".

Input

The first line contains two integers n and w (1?≤?n,?w?≤?2·105) ? the number of towers in the bears' and the elephant's walls correspondingly. The second line contains n integers ai (1?≤?ai?≤?109) ? the heights of the towers in the bears' wall. The third line contains w integers bi (1?≤?bi?≤?109) ? the heights of the towers in the elephant's wall.

Output

Print the number of segments in the bears' wall where Horace can "see an elephant".

Sample test(s)

input

13 52 4 5 5 4 3 2 2 2 3 3 2 13 4 4 3 2

output

Note

The picture to the left shows Horace's wall from the sample, the picture to the right shows the bears' wall. The segments where Horace can "see an elephant" are in gray.

预处理出差值跑KMP.....

才发现我的KMP模版有一个小bug.....还用了这么多年......

/** * Created by ckboss on 14-9-27. */import java.util.*;import java.io.*;public class CF471D {    static int n,m;    static int[] hi = new int[200200];    static int[] wi = new int[200200];    static int[] h = new int[200200];    static int[] w = new int[200200];    static int[] f = new int[200200];    static void getfail(){        f[0]=f[1]=0;        for(int i=1;i<m;i++){            int j=f[i];            while(j!=0&&w[j]!=w[i]) j=f[j];            f[i+1]=(w[i]==w[j])?j+1:0;        }    }    static int kmp(){        int ans=0;        getfail();        int j=0;        for(int i=0;i<n;i++){            while(j!=0&&w[j]!=h[i]) j=f[j];            if(w[j]==h[i]) j++;            if(j==m){                ans++;                j=f[j];            }        }        return ans;    }    public static void main(String[] args){        Scanner in = new Scanner(System.in);        n=in.nextInt(); m=in.nextInt();        for(int i=0;i<n;i++){            hi[i]=in.nextInt();            if(i-1>=0) h[i-1]=hi[i]-hi[i-1];        }        for(int i=0;i<m;i++) {            wi[i] = in.nextInt();            if(i-1>=0) w[i-1]=wi[i]-wi[i-1];        }        m--;n--;        if(m==0){            System.out.println(n+1);            return ;        }        System.out.println(kmp());    }}