catalogue
2. Hexadecimal to octal number
5. Number of special palindromes
1. Sequence sorting
Given a sequence of length N, the sequence is arranged from small to large. 1<=n<=200
Input format
The first line is an integer n.
The second row contains n integers, which are the numbers to be sorted, and the absolute value of each integer is less than 10000.
Output format
Output a row, and output the sorted sequence from small to large.
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in= new Scanner(System.in);
int n=in.nextInt();
int []a=new int[n];
for(int i=0;i<n;i++) {
a[i]=in.nextInt();
}
Arrays.sort(a);
for(int i =0;i<n;i++) {
System.out.printf("%d ",a[i]);
}
}
}
2. Hexadecimal to octal number
Given n hexadecimal positive integers, output their corresponding octal numbers.
Input format
The first line of input is a positive integer n (1 < = n < = 10).
In the next n lines, a string composed of 0 ~ 9 and uppercase letters A~F in each line represents the hexadecimal positive integer to be converted, and the length of each hexadecimal number shall not exceed 100000.
Output format
Output n lines, and input the corresponding octal positive integer for each line.
[note]
The entered hexadecimal number will not have a leading 0, such as 012A.
The output octal number cannot have a leading 0.
import java.math.BigInteger;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in =new Scanner(System.in);
int n=in.nextInt();
String []a = new String[n];
for(int i=0;i<n;i++){
a[i]=in.next();
}
for(int i=0;i<n;i++){
System.out.println(new BigInteger(a[i],16).toString(8));
}
}
}
3. Hexadecimal to decimal
Input a positive hexadecimal digit string of no more than 8 digits from the keyboard, convert it into a positive decimal number and output it.
Note: 10 ~ 15 in hexadecimal numbers are represented by uppercase English letters A, B, C, D, E and F respectively.
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
String a=sc.next();
System.out.print(new BigInteger(a,16).toString(10));
}
}
4. Decimal to hexadecimal
Hexadecimal number is an integer representation often used in programming. It has 16 symbols of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a, B, C, D, e and F, representing 0 to 15 decimal numbers respectively. The hexadecimal counting method is full 16 into 1, so the decimal number 16 is 10 in hexadecimal, while the decimal number 17 is 11 in hexadecimal, and so on. The decimal number 30 is 1E in hexadecimal. Give a nonnegative integer and express it in hexadecimal form.
Input format
The input contains a nonnegative integer a representing the number to convert. 0<=a<=2147483647
Output format
Outputs the hexadecimal representation of this integer
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String a = sc.next();
System.out.print(new BigInteger(a,10).toString(16).toUpperCase());
}
}import java.util.*;
import java.math.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int a = sc.nextInt();
System.out.print(Integer.toHexString(a).toUpperCase());
}
}
5. Number of special palindromes
123321 is a very special number. It reads from the left and from the right.
Enter a positive integer n, and program to find all such five and six digit decimal numbers, so that the sum of each number is equal to n.
Input format
Enter a line containing a positive integer n.
Output format
Output qualified integers from small to large, and each integer occupies one line.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
for(int i = 10000;i<=99999;i++) {
int a = i/10000%10;
int b = i/1000%10;
int c = i/100%10;
int d = i/10%10;
int e = i%10;
if(a==e&&b==d&&a+b+c+d+e==n) {
System.out.print(i+"\n");
}
}
for(int i = 100000;i<=999999;i++) {
int f = i/100000%10;
int a = i/10000%10;
int b = i/1000%10;
int c = i/100%10;
int d = i/10%10;
int e = i%10;
if(f==e&&a==d&&b==c&&a+b+c+d+e+f==n) {
System.out.print(i+"\n");
}
}
}
}
6. Palindromes
1221 is a very special number. It reads from the left and from the right. It is programmed to find all such four digit decimal numbers.
Output format
Output four decimal numbers that meet the conditions in the order from small to large.
public class Main {
public static void main(String[] args) {
int i;
for(i=0;i<10000;i++) {
if(i>1000) {
int a = i/1000%10;
int b = i/100%10;
int c = i/10%10;
int d = i%10;
if(a==d&&b==c) {
System.out.print(i+"\n");
}
}
}
}
}
7. Special numbers
Problem description
153 is a very special number, which is equal to the cubic sum of each number, that is, 153 = 1 * 1 * 1 + 5 * 5 * 5 + 3 * 3 * 3. Programming to find all three decimal numbers that meet this condition.
Output format
Output three decimal numbers that meet the conditions from small to large, and each number occupies one line.
public class Main {
public static void main(String[] args) {
int i;
for(i=0;i<1000;i++) {
if(i>99) {
int a = i/100%10;
int b = i/10%10;
int c = i%10;
if(a*a*a+b*b*b+c*c*c==i) {
System.out.print(i+"\n");
}
}
}
}
}
8. Yang Hui triangle
Problem description
Yang Hui triangle is also called Pascal triangle. Its i+1 line is the coefficient of the expansion of (a+b)i.
One of its important properties is that each number in a triangle is equal to the sum of the numbers on its two shoulders
The first four rows of Yang Hui triangle are given below:
1
1 1
1 2 1
1 3 3 1
Give n and output its first n lines.
Input format
The input contains a number n.
Output format
Output the first n rows of Yang Hui triangle. Each line is output from the first number of this line, separated by a space. Please do not output extra spaces in front.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[][] = new int [n][n];
for(int i=0;i<n;i++) {
a[i][0]=1;//Each line begins with a 1
a[i][i]=1;//The value in row i and column i is 1
}
for(int i=2;i<n;i++) {
for(int j=1;j<=i-1;j++) {
a[i][j]=a[i-1][j-1]+a[i-1][j];
}
}
for(int i=0;i<n;i++) {
for(int j=0;j<=i;j++) {
System.out.print(a[i][j]+" ");
}
System.out.println();
}
}
}
9. Find integer
Problem description
Given a sequence containing n integers, ask the first occurrence of integer a in the sequence.
Input format
The first line contains an integer n.
The second row contains n non negative integers, which is a given sequence, and each number in the sequence is not greater than 10000.
The third line contains an integer a, which is the number to be found.
Output format
If a appears in the sequence, output the position where it first appears (the position is numbered from 1), otherwise output - 1.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int []a=new int[n];
for(int i=0;i<n;i++) {
a[i]=sc.nextInt();
}
int b = sc.nextInt();
Boolean find = false;
for(int i=0;i<n;i++) {
if(b==a[i]) {
find=true;
System.out.print(i+1);
break;
}
}
if(!find) {
System.out.println(-1);
}
}
}
10. Sequence characteristics
Problem description
Give n numbers and find the maximum, minimum, and.
Input format
The first line is an integer n, which represents the number of numbers.
The second line has n numbers, which are given n numbers, and the absolute value of each number is less than 10000.
Output format
Output three lines, one integer per line. The first row represents the maximum of these numbers, the second row represents the minimum of these numbers, and the third row represents the sum of these numbers.
import java.lang.reflect.Array;
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] a = new int[n];
for(int i=0;i<n;i++) {
a[i]=sc.nextInt();
}
int max=0;
int min=0;
int sum=0;
Arrays.sort(a);
max=a[n-1];
min=a[0];
for(int i=0;i<n;i++) {
sum=sum+a[i];
}
System.out.print(max+"\n");
System.out.print(min+"\n");
System.out.print(sum+"\n");
}
}
11. Letter graphics
Problem description
Some beautiful figures can be formed by using letters. An example is given below:
ABCDEFG
BABCDEF
CBABCDE
DCBABCD
EDCBABC
This is a graph with 5 rows and 7 columns. Please find out the law of this graph and output a graph with n rows and m columns.
Input format
Enter a row containing two integers n and m, which respectively represent the number of rows and columns of the graph you want to output.
Output format
Output n lines, each m characters, for your graphics.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
for(int i=0;i<n;i++){
for(int j=i;j>=0 && i-j<m;j--){
System.out.print((char)(65+j));
}
for(int k=1;k<m-i;k++){
System.out.print((char)(65+k));
}
System.out.println();
}
}
}
12, 01 string
Problem description
For a 01 string with a length of 5 bits, each bit may be 0 or 1. There are 32 possibilities in total. The first few of them are:
00000
00001
00010
00011
00100
Please output these 32 01 strings in descending order.
Output format
Output 32 lines, in order from small to large, each line has a 01 string with a length of 5.
public class Main {
public static void main(String[] args) {
for(int i=0;i<32;i++)
{
int j=i+256;
String a = Integer.toBinaryString(j);
String c=a.substring(a.length()-5);
System.out.print(c+"\n");
}
}
}
13. Leap year judgment
Problem description
Given a year, judge whether it is a leap year.
This year is a leap year when one of the following conditions is met:
1. The year is a multiple of 4 rather than a multiple of 100;
2. The year is a multiple of 400.
Other years are not leap years.
Input format
The input contains an integer y representing the current year.
Output format
Output a line. If the given year is a leap year, output yes, otherwise output no.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int year = sc.nextInt();
if(year%4==0 && year%100!=0 || year%400==0) {
System.out.print("yes");
}else {
System.out.print("no");
}
}
}
14. Fibonacci sequence
Problem description
The recurrence formula of Fibonacci sequence is: Fn=Fn-1+Fn-2, where F1=F2=1.
When n is large, Fn is also very large. Now we want to know what is the remainder of Fn divided by 10007.
Input format
The input contains an integer n.
Output format
Output a line containing an integer representing the remainder of Fn divided by 10007.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] f = new int[n+2];
f[1]=1;f[2]=1;
if(n>2) {
for(int i = 3;i<=n;i++) {
f[i]=(f[i-1]+f[i-2])%10007;
}
}
System.out.print(f[n]);
}
}
15. Area of circle
Problem description
Given the radius r of a circle, find the area of the circle.
Input format
The input contains an integer r representing the radius of the circle.
Output format
Output a line containing a real number, rounded to 7 digits after the decimal point, indicating the area of the circle.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
//PI=3.14159265358979323
final double PI=3.14159265358979323;
Scanner sc = new Scanner(System.in);
int r = sc.nextInt();
double s = PI*r*r;
System.out.printf("%.7f",s);
}
}
16. Sequence summation
Problem description
Find 1 + 2 + 3 ++ The value of n.
Input format
The input includes an integer n.
Output format
Output a line, including an integer, representing 1 + 2 + 3 ++ The value of n.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc =new Scanner(System.in);
int n = sc.nextInt();
long sum = 0;
for(int i=1;i<=n;i++) {
sum = sum+i;
}
System.out.println(sum);
}
}
