AtCoder Beginner Contest 106 2018/08/18

Problem Statement

There is a farm whose length and width are A yard and B yard, respectively. A farmer, John, made a vertical road and a horizontal road inside the farm from one border to another, as shown below: (The gray part represents the roads.)

What is the area of this yard excluding the roads? Find it.

Note

It can be proved that the positions of the roads do not affect the area.

Constraints

A is an integer between 2 and 100 (inclusive).
B is an integer between 2 and 100 (inclusive).

Sample Input 1

Copy

2 2

Sample Output 1

Copy

In this case, the area is 1 square yard.

Sample Input 2

Copy

5 7

Sample Output 2

Copy

In this case, the area is 24 square yards.

容斥定理。

代码：

import java.util.*;

public class Main {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int a = in.nextInt();
        int b = in.nextInt();
        System.out.println(a * b - a - b + 1);
    }
}

B - 105

Time limit : 2sec / Memory limit : 1000MB

Score: 200 points

Problem Statement

The number 105 is quite special - it is odd but still it has eight divisors. Now, your task is this: how many odd numbers with exactly eight positive divisors are there between 1 and N (inclusive)?

Constraints

N is an integer between 1 and 200 (inclusive).

Input

Input is given from Standard Input in the following format:

Output

Print the count.

Sample Input 1

Copy

Sample Output 1

Copy

Among the numbers between 1 and 105, the only number that is odd and has exactly eight divisors is 105.

Sample Input 2

Copy

Sample Output 2

Copy

1 has one divisor. 3, 5 and 7 are all prime and have two divisors. Thus, there is no number that satisfies the condition.

代码：

import java.util.*;

public class Main {
    static int get(int a) {
        int b = 2;
        for(int i = 3;i <= a / 3;i += 2) {
            if(a % i == 0) {
                b ++;
            }
        }
        return b;
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int a = in.nextInt();
        int m = 0;
        if(a >= 105)m ++;
        for(int i = 107;i <= a;i += 2) {
            if(get(i) == 8)m ++;
        }
        System.out.println(m);
    }
}

C - To Infinity

Time limit : 2sec / Memory limit : 1000MB

Score: 300 points

Problem Statement

Mr. Infinity has a string S consisting of digits from 1 to 9. Each time the date changes, this string changes as follows:

Each occurrence of 2 in S is replaced with 22. Similarly, each 3 becomes 333, 4 becomes 4444, 5 becomes 55555, 6 becomes 666666, 7 becomes 7777777, 8 becomes 88888888 and 9 becomes 999999999. 1 remains as 1.

For example, if S is 1324, it becomes 1333224444 the next day, and it becomes 133333333322224444444444444444 the day after next. You are interested in what the string looks like after 5×1015 days. What is the K-th character from the left in the string after 5×1015 days?

Constraints

S is a string of length between 1 and 100 (inclusive).
K is an integer between 1 and 1018 (inclusive).
The length of the string after 5×1015 days is at least K.

Input

Input is given from Standard Input in the following format:

S
K

Output

Print the K-th character from the left in Mr. Infinity's string after 5×1015 days.

Sample Input 1

Copy

1214
4

Sample Output 1

Copy

The string S changes as follows:

Now: 1214
After one day: 12214444
After two days: 1222214444444444444444
After three days: 12222222214444444444444444444444444444444444444444444444444444444444444444

The first five characters in the string after 5×1015 days is 12222. As K=4, we should print the fourth character, 2.

Sample Input 2

Copy

3
157

Sample Output 2

Copy

The initial string is 3. The string after 5×1015 days consists only of 3.

Sample Input 3

Copy

299792458
9460730472580800

Sample Output 3

Copy

2
只需要看看开头有几个连续的1即可。
代码：

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String s = in.nextLine();
        long a = in.nextLong();
        int b = 0;
        for(int i = 0;i < s.length();i ++) {
            if(s.charAt(i) == '1')b ++;
            else break;
        }
        if(b >= a)System.out.println(1);
        else System.out.println(s.charAt(b));
    }
}