BestCoder Round #11 (Div. 2) 题解

2022-03-27 00:13:46

HDOJ5054 Alice and Bob

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 302    Accepted Submission(s): 229

Problem Description
Bob and Alice got separated in the Square, they agreed that if they get separated, they'll meet back at the coordinate point (x, y). Unfortunately they forgot to define the origin of coordinates and the coordinate axis direction. Now, Bob in the lower left
corner of the Square, Alice in the upper right corner of the the Square. Bob regards the lower left corner as the origin of coordinates, rightward for positive direction of axis X, upward for positive direction of axis Y. Alice regards the upper right corner
as the origin of coordinates, leftward for positive direction of axis X, downward for positive direction of axis Y. Assuming that Square is a rectangular, length and width size is N * M. As shown in the figure:

BestCoder Round #11 (Div. 2) 题解


Bob and Alice with their own definition of the coordinate system respectively, went to the coordinate point (x, y). Can they meet with each other ? 

Note: Bob and Alice before reaching its destination, can not see each other because of some factors (such as buildings, time poor).
 
Input
There are multiple test cases. Please process till EOF. Each test case only contains four integers : N, M and x, y. The Square size is N * M, and meet in coordinate point (x, y). ( 0 < x < N <= 1000 , 0 < y < M <= 1000 ).
 
Output
If they can meet with each other, please output "YES". Otherwise, please output "NO".
 
Sample Input
10 10 5 5

10 10 6 6
 
Sample Output
YES

NO
 
Source
 

/**

 * Created by ckboss on 14-10-3.

 */

import java.util.*;

public class Main {

    public static void main(String[] args){

        Scanner in = new Scanner(System.in);

        while(in.hasNext())

        {

            int N=in.nextInt();

            int M=in.nextInt();

            int x=in.nextInt();

            int y=in.nextInt();

            if(x==N-x&&y==M-y)

                System.out.println("YES");

            else

                System.out.println("NO");

        }

    }

}

HDOJ5055 Bob and math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 812    Accepted Submission(s): 313

Problem Description
Recently, Bob has been thinking about a math problem.

There are N Digits, each digit is between 0 and 9. You need to use this N Digits to constitute an Integer.

This Integer needs to satisfy the following conditions:
  • 1. must be an odd Integer.
  • 2. there is no leading zero.
  • 3. find the biggest one which is satisfied 1, 2.

Example: 

There are three Digits: 0, 1, 3. It can constitute six number of Integers. Only "301", "103" is legal, while "130", "310", "013", "031" is illegal. The biggest one of odd Integer is "301".

 
Input
There are multiple test cases. Please process till EOF.

Each case starts with a line containing an integer N ( 1 <= N <= 100 ).

The second line contains N Digits which indicate the digit $a_1, a_2, a_3, \cdots, a_n. ( 0 \leq a_i \leq 9)$.
 
Output
The output of each test case of a line. If you can constitute an Integer which is satisfied above conditions, please output the biggest one. Otherwise, output "-1" instead.
 
Sample Input
3

0 1 3

3

5 4 2

3

2 4 6
 
Sample Output
301

425

-1
 
Source
 

/**

 * Created by ckboss on 14-10-3.

 */

import java.util.*;

public class Main {

    static int[] num = new int[10];

    static char[] ans = new char[110];

    static int nt;

    public static void main(String[] args){

        Scanner in = new Scanner(System.in);

        int n;

        while(in.hasNext()){

            n=in.nextInt();

            int flag=0;

            int MinJ=111;

            Arrays.fill(num,0); nt=0;

            for(int i=0;i<n;i++){

                int x=in.nextInt();

                num[x]++;

                if(x%2==1){

                    if(x<MinJ) MinJ=x;

                    flag++;

                }

            }

            if(flag==0)

            {

                System.out.println("-1");

                continue;

            }

            num[MinJ]--;

            for(int i=9;i>=0;i--){

                for(int j=num[i];j>0;j--){

                    ans[nt++]=(char)('0'+i);

                }

            }

            ans[nt++]=(char)(MinJ+'0');

            flag=0;

            for(int i=0;i<nt;i++){

                if(ans[i]=='0' && flag==0){

                    flag=0; break;

                }

                flag=1;

                System.out.print(ans[i]);

            }

            if(flag==1)

                System.out.println("");

            else

                System.out.println("-1");

        }

    }

}

HDOJ5056 Boring count

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 615    Accepted Submission(s): 242

Problem Description
You are given a string S consisting of lowercase letters, and your task is counting the number of substring that the number of each lowercase letter in the substring is no more than K.
 
Input
In the first line there is an integer T , indicates the number of test cases.

For each case, the first line contains a string which only consist of lowercase letters. The second line contains an integer K.



[Technical Specification]

1<=T<= 100

1 <= the length of S <= 100000

1 <= K <= 100000
 
Output
For each case, output a line contains the answer.
 
Sample Input
3

abc

1

abcabc

1

abcabc

2
 
Sample Output
6

15

21
 
Source
 

O(n)贪心...

/**

 * Created by ckboss on 14-10-3.

 */

import java.util.*;

public class Main {

    static long[] num = new long[30];

    public static void main(String[] args){

        Scanner in = new Scanner(System.in);

        int T_T=in.nextInt();

        while(T_T-->0){

            String st=in.next();

            int k=in.nextInt();

            Arrays.fill(num,0);

            int be=0,ed=0;

            long ans=0;

            for(int i=0,sz=st.length();i<sz;i++){

                int id=(int)(st.charAt(i)-'a');

                num[id]++;

                if(num[id]>k){

                    for(;be<=ed;be++){

                        ans+=ed-be+1;

                        num[st.charAt(be)-'a']--;

                        if(st.charAt(be)-'a'==id) {

                            be++;

                            break;

                        }

                    }

                }

                ed=i;

            }

            for(;be<=ed;be++){

                ans+=ed-be+1;

                num[st.charAt(be)-'a']--;

            }

            System.out.println(ans);

        }

    }

}

HDOJ5057 Argestes and Sequence

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 603    Accepted Submission(s): 152

Problem Description
Argestes has a lot of hobbies and likes solving query problems especially. One day Argestes came up with such a problem. You are given a sequence a consisting of N nonnegative integers, a[1],a[2],...,a[n].Then there are M operation on the sequence.An operation
can be one of the following:

S X Y: you should set the value of a[x] to y(in other words perform an assignment a[x]=y).

Q L R D P: among [L, R], L and R are the index of the sequence, how many numbers that the Dth digit of the numbers is P.

Note: The 1st digit of a number is the least significant digit.
 
Input
In the first line there is an integer T , indicates the number of test cases.

For each case, the first line contains two numbers N and M.The second line contains N integers, separated by space: a[1],a[2],...,a[n]—initial value of array elements.

Each of the next M lines begins with a character type.

If type==S,there will be two integers more in the line: X,Y.

If type==Q,there will be four integers more in the line: L R D P.



[Technical Specification]

1<=T<= 50

1<=N, M<=100000

0<=a[i]<=$2^{31}$ - 1

1<=X<=N

0<=Y<=$2^{31}$ - 1

1<=L<=R<=N

1<=D<=10

0<=P<=9
 
Output
For each operation Q, output a line contains the answer.
 
Sample Input
1

5 7

10 11 12 13 14

Q 1 5 2 1

Q 1 5 1 0

Q 1 5 1 1

Q 1 5 3 0

Q 1 5 3 1

S 1 100

Q 1 5 3 1
 
Sample Output
5

1

1

5

0

1
 
Source
 

分块大法好....

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

const int maxn=100100;

struct BLOCK

{

    int cnt[10][10];

}block[400];

int block_size,block_num;

int n,m;

int a[maxn];

const int ten[12]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};

void CHANGE(int p,int v)

{

    int id=p/block_size;

    int x=a[p];

    for(int i=0;i<10;i++)

    {

        block[id].cnt[i][x%10]--;

        x/=10;

    }

    a[p]=v;

    x=v;

    for(int i=0;i<10;i++)

    {

        block[id].cnt[i][x%10]++;

        x/=10;

    }

}

int QUERY(int l,int r,int p,int d)

{

    int L=l/block_size,R=r/block_size;

    int ans=0;

    if(R-L<=1)

    {

        for(int i=l;i<=r;i++)

        {

            ans+=((a[i]/ten[p-1])%10==d)?1:0;

        }

        return ans;

    }

    for(int i=l;i<(L+1)*block_size;i++)

    {

        ans+=((a[i]/ten[p-1])%10==d)?1:0;

    }

    for(int i=L+1;i<=R-1;i++)

    {

        ans+=block[i].cnt[p-1][d];

    }

    for(int i=R*block_size;i<=r;i++)

    {

        ans+=((a[i]/ten[p-1])%10==d)?1:0;

    }

    return ans;

}

int main()

{

    int T_T;

    scanf("%d",&T_T);

    while(T_T--)

    {

        scanf("%d%d",&n,&m);

        memset(block,0,sizeof(block));

        block_size=sqrt(n*1.0)+1;

        block_num=n/block_size+1;

        for(int i=1;i<=n;i++)

        {

            scanf("%d",&a[i]);

            int x=a[i];

            int id=i/block_size;

            for(int j=0;j<10;j++)

            {

                block[id].cnt[j][x%10]++;

                x/=10;

            }

        }

        while(m--)

        {

            char op[10];

            int a,b,c,d;

            scanf("%s",op);

            if(op[0]=='Q')

            {

                scanf("%d%d%d%d",&a,&b,&c,&d);

                printf("%d\n",QUERY(a,b,c,d));

            }

            else if(op[0]=='S')

            {

                scanf("%d%d",&a,&b);

                CHANGE(a,b);

            }

        }

    }

    return 0;

}
上一篇：[Tex学习]WinEdit 常用软件快捷键


下一篇：从CUMT校园导航出现的问题看CSS布局设计（一） CSS盒模型