AtCoder Beginner Contest 096

A - Day of Takahashi

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Time limit : 2sec / Memory limit : 256MB

Score: 100 points

Problem Statement

In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year.
For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year.

In this country, a date is called Takahashi when the month and the day are equal as numbers. For example, 5-5 is Takahashi.
How many days from 2018-1-1 through 2018-a-b are Takahashi?

Constraints

• a is an integer between 1 and 12 (inclusive).
• b is an integer between 1 and 31 (inclusive).
• 2018-a-b is a valid date in Gregorian calendar.

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Input

Input is given from Standard Input in the following format:

a b

Output

Print the number of days from 2018-1-1 through 2018-a-b that are Takahashi.

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Sample Input 1

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5 5

Sample Output 1

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5

There are five days that are Takahashi: 1-1, 2-2, 3-3, 4-4 and 5-5.

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Sample Input 2

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2 1

Sample Output 2

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1

There is only one day that is Takahashi: 1-1.

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Sample Input 3

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11 30

Sample Output 3

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11

There are eleven days that are Takahashi: 1-1, 2-2, 3-3, 4-4, 5-5, 6-6, 7-7, 8-8, 9-9, 10-10 and 11-11.

#include <iostream>
 
using namespace std;
 
int main(){
    int a,b;
    cin>>a>>b;
    cout<<(b>=a?a:(a-1))<<endl;
    return 0;
}

B - Maximum Sum

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Time limit : 2sec / Memory limit : 256MB

Score: 200 points

Problem Statement

There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times:

• Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n.

What is the largest possible sum of the integers written on the blackboard after K operations?

Constraints

• A,B and C are integers between 1 and 50 (inclusive).
• K is an integer between 1 and 10 (inclusive).

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Input

Input is given from Standard Input in the following format:

A B C
K

Output

Print the largest possible sum of the integers written on the blackboard after K operations by E869220.

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Sample Input 1

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5 3 11
1

Sample Output 1

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30

In this sample, 5,3,11 are initially written on the blackboard, and E869120 can perform the operation once.
There are three choices:

1. Double 5: The integers written on the board after the operation are 10,3,11.
2. Double 3: The integers written on the board after the operation are 5,6,11.
3. Double 11: The integers written on the board after the operation are 5,3,22.

If he chooses 3., the sum of the integers written on the board afterwards is 5+3+22=30, which is the largest among 1. through 3.

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Sample Input 2

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3 3 4
2

Sample Output 2

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22

E869120 can perform the operation twice. The sum of the integers eventually written on the blackboard is maximized as follows:

• First, double 4. The integers written on the board are now 3,3,8.
• Next, double 8. The integers written on the board are now 3,3,16.

Then, the sum of the integers eventually written on the blackboard is 3+3+16=22.

#include <iostream>
#define ll long long int
using namespace std;
 
int main(){
    ll a,b,c,k;
    cin>>a>>b>>c>>k;
    ll x = max(c,max(a,b));
    ll shit = a + b + c - x;
    while(k--){
        x<<=1;
    }
    cout<<x+shit<<endl;
    return 0;
}

C - Grid Repainting 2

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Time limit : 2sec / Memory limit : 256MB

Score: 300 points

Problem Statement

We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i,j).
Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i,j) black when si,j= #, and to make Square (i,j) white when si,j= ..
However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black.
Determine if square1001 can achieve his objective.

Constraints

• H is an integer between 1 and 50 (inclusive).
• W is an integer between 1 and 50 (inclusive).
• For every (i,j) (1≤i≤H,1≤j≤W), si,j is # or ..

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Input

Input is given from Standard Input in the following format:

H W
s1,1s1,2s1,3…s1,W
s2,1s2,2s2,3…s2,W
  :   :
sH,1sH,2sH,3…sH,W

Output

If square1001 can achieve his objective, print Yes; if he cannot, print No.

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Sample Input 1

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3 3
.#.
###
.#.

Sample Output 1

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Yes

One possible way to achieve the objective is shown in the figure below. Here, the squares being painted are marked by stars.

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Sample Input 2

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5 5
#.#.#
.#.#.
#.#.#
.#.#.
#.#.#

Sample Output 2

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No

square1001 cannot achieve his objective here.

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Sample Input 3

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11 11
...#####...
.##.....##.
#..##.##..#
#..##.##..#
#.........#
#...###...#
.#########.
.#.#.#.#.#.
##.#.#.#.##
..##.#.##..
.##..#..##.

Sample Output 3

Copy

Yes

 

#include <iostream>
#define ll long long int
#define N 55
using namespace std;
char s[N][N];
int dir[4][2]={0,1,1,0,0,-1,-1,0};
int a,b;
 
bool search(int x,int y){
    for(int i=0;i<4;i++){
        int xx = x+dir[i][0];
        int yy = y+dir[i][1];
        if(xx>=0&&xx<a&&y>=0&&y<b){
            if(s[xx][yy]=='#')
                return true;
        }
    }
    return false;
}
int main(){
    cin>>a>>b;
    for(int i=0;i<a;i++)
        for(int j=0;j<b;j++)
            cin>>s[i][j];
    for(int i=0;i<a;i++){
        for(int j=0;j<b;j++){
            if(s[i][j]=='#'){
                bool prime = search(i,j);
                if(!prime){
                    return 0*printf("No
");
                }
            }
        }
    }
    printf("Yes
");
    return 0;
}

D - Five, Five Everywhere

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Time limit : 2sec / Memory limit : 256MB

Score: 400 points

Problem Statement

Print a sequence a1,a2,…,aN whose length is N that satisfies the following conditions:

• ai (1≤i≤N) is a prime number at most 55 555.
• The values of a1,a2,…,aN are all different.
• In every choice of five different integers from a1,a2,…,aN, the sum of those integers is a composite number.

If there are multiple such sequences, printing any of them is accepted.

Notes

An integer N not less than 2 is called a prime number if it cannot be divided evenly by any integers except 1 and N, and called a composite number otherwise.

Constraints

• N is an integer between 5 and 55 (inclusive).

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Input

Input is given from Standard Input in the following format:

N

Output

Print N numbers a1,a2,a3,…,aN in a line, with spaces in between.

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Sample Input 1

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5

Sample Output 1

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3 5 7 11 31

Let us see if this output actually satisfies the conditions.
First, 3, 5, 7, 11 and 31 are all different, and all of them are prime numbers.
The only way to choose five among them is to choose all of them, whose sum is a1+a2+a3+a4+a5=57, which is a composite number.
There are also other possible outputs, such as 2 3 5 7 13, 11 13 17 19 31 and 7 11 5 31 3.

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Sample Input 2

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6

Sample Output 2

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2 3 5 7 11 13

• 2, 3, 5, 7, 11, 13 are all different prime numbers.
• 2+3+5+7+11=28 is a composite number.
• 2+3+5+7+13=30 is a composite number.
• 2+3+5+11+13=34 is a composite number.
• 2+3+7+11+13=36 is a composite number.
• 2+5+7+11+13=38 is a composite number.
• 3+5+7+11+13=39 is a composite number.

Thus, the sequence 2 3 5 7 11 13 satisfies the conditions.

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Sample Input 3

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8

Sample Output 3

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2 5 7 13 19 37 67 79

要想5个数任意一个都可以被整除，不如就规定好被5整除，那么每个数只要n×5+1就行了。

#include <iostream>
using namespace std;
int an[55566];
void search() {
    an[0] = 1,an[1] = 1;
    for(int i=2;i<=55555;i++){
        if(!an[i]){
            for(int j=2;j*i<=55555;j++){
                an[i*j] = 1;
            }
        }
    }
}
int main() {
  int n, count = 2;
  search();
  cin >> n;
  int prime[n];
  for (int i = 0; i < n;) {
    if ( !an[count] && count % 5 == 1) {
      prime[i] = count;
      i++;
    }
    count++;
  }
  for (int i = 0; i < n; i++) {
    cout << prime[i] <<" ";
  }
  cout<<endl;
  return 0;
}
#include <iostream>
using namespace std;
int an[55566];
void search() {
    an[0] = 1,an[1] = 1;
    for(int i=2;i<=55555;i++){
        if(!an[i]){
            for(int j=2;j*i<=55555;j++){
                an[i*j] = 1;
            }
        }
    }
}
int main() {
  int n, count = 2;
  search();
  cin >> n;
  int prime[n];
  for (int i = 0; i < n;) {
    if ( !an[count] && count % 5 == 1) {
      prime[i] = count;
      i++;
    }
    count++;
  }
  for (int i = 0; i < n; i++) {
    cout << prime[i] <<" ";
  }
  cout<<endl;
  return 0;
}