Luogu------Fundamentals 1035-Summation of Series 1980-Counting Problems

1035 topic description

Known: Sn​=1+1/2​+1/3​+…+1/n. Obviously for any integer k, when n is large enough, S_n>k.

Given an integer k, it is required to calculate a minimum n such that S_n>k.

input format

A positive integer k.

output format

A positive integer n.

Input and output samples

enter

1

output

2

Instructions/Tips

【data range】

For 100% of the data, 1≤k≤15.

 

analyze:

Enter a positive integer K

S_n = 1+ 1/2 + 1/3 +...+1/n;

such that S_n>K    

output a positive integer n

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        double sum=1;

        int k = sc.nextInt();

        for (int i = 2; i <= 9999999 ; i++) {
            sum+=(double)1/i;

            if(sum>k){
                System.out.println(i);
                break;
            }

        }

    }
}

 

 

1980 Title Description

Try to count how many times the number x appears among all integers in the interval 1 to n? For example, in 1 to 11, that is, in 1,2,3,4,5,6,7,8,9,10,11, the number 11 occurs 4 times.

input format

2 integers n and x are separated by a space.

output format

1 integer representing the number of occurrences of x.

Input and output samples

enter 

11 1

output 

4

Instructions/Tips

For 100\%100% of the data 1≤n≤10^6

 

analyze:

Here, it is only necessary to judge the remainder of each judgment to judge the value of the single digit.

  y=i;
            while(y>0){
                if(y%10==x) {
                    sum++;
                }
                y/=10;
            }

import java.util.Scanner;


public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int sum=0;
        int y;

        int n = sc.nextInt();
        int x = sc.nextInt();

        for (int i = 1; i <=n ; i++) {
            y=i;
            while(y>0){
                if(y%10==x) {
                    sum++;
                }
                y/=10;
            }

        }
        System.out.println(sum);
    }

}