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2026题:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
while (sc_01.hasNext()) {
String strings = sc_01.nextLine();
char[] output = new char[strings.length()];
for (int i = 0; i < strings.length(); i++) {
if (i == 0) {
if (strings.charAt(0) <= 122 && strings.charAt(0) >= 97) {
output[0] = (char) (strings.charAt(0) - 32);
} else {
output[0] = strings.charAt(0);
}
} else {
if (strings.charAt(i-1) == 32 && strings.charAt(i) <= 122 && strings.charAt(i) >= 97) {
output[i] = (char) (strings.charAt(i) - 32);
} else {
output[i] = strings.charAt(i);
}
}
}
for (int i = 0; i < strings.length(); i++) {
System.out.print(output[i]);
}
System.out.println();
}
}
}
2027题:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
int num = sc_01.nextInt();
sc_01.nextLine();
for (int j = 0; j < num; j++) {
String strings = sc_01.nextLine();
int[] yuan_yin = new int[5];
//计算每一个元音的数量
for (int i = 0; i < strings.length(); i++) {
char char_vol = strings.charAt(i);
switch (char_vol) {
case 'a':
yuan_yin[0]++;
break;
case 'e':
yuan_yin[1]++;
break;
case 'i':
yuan_yin[2]++;
break;
case 'o':
yuan_yin[3]++;
break;
case 'u':
yuan_yin[4]++;
break;
default:
break;
}
}
//对统计结果进行输出
for (int i = 0; i < yuan_yin.length; i++) {
switch (i) {
case 0:
System.out.println("a:" + yuan_yin[0]);
break;
case 1:
System.out.println("e:" + yuan_yin[1]);
break;
case 2:
System.out.println("i:" + yuan_yin[2]);
break;
case 3:
System.out.println("o:" + yuan_yin[3]);
break;
case 4:
System.out.println("u:" + yuan_yin[4]);
break;
default:
break;
}
}
if (j < num - 1) {
System.out.println();
} else {
}
}
}
}
2028题:
import java.util.*;
public class Main {
//(1)求最大公约数法:最大公约数*最小公倍数 = 两个数之积。
//(2)使用辗转相除法求 比如:下图为求 2 4 6的最小公倍数,用2 4 6做辗转相除法可得最小公倍数为2*1*2
// 3.而第三种方法为:先求出n个数中最大的那个数max,如果这个数可以整除所有的数,则这是最小公倍数
// 如果不能则令max+1直到找到可以整除所有的数的那个数为止
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
while (sc_01.hasNext()) {
int char_num = sc_01.nextInt();
int[] all_num = new int[char_num];
for (int i = 0; i < char_num; i++) {
all_num[i] = sc_01.nextInt();
}
int counting = 0;
//计算最大公约数
int Max = check(all_num, counting);
//计算最小公倍数
int result = lease_multiple(all_num, counting, Max);
all_num[counting + 1] = result;
//循环计算两个数的最小公倍数,其中一个是之前两个数的公倍数。
while (counting < char_num - 2) {
counting++;
Max = check(all_num, counting);
result = lease_multiple(all_num, counting, Max);
all_num[counting+1] = result;
}
System.out.println(result);
}
}
private static int lease_multiple(int[] all_num, int counting, int max) {
int result = (all_num[counting] / max) * all_num[counting + 1];
return result;
}
private static int check(int[] all_num, int counting) {
//先找出其中的最小值。
int min = all_num[counting];
if (all_num[counting + 1] < min) {
int tmp = all_num[counting];
all_num[counting] = all_num[counting + 1];
all_num[counting + 1] = tmp;
}
int flag = 0;
//反向遍历获取最大公约数。
for (int i = all_num[counting]; i >= 1; i--) {
if (all_num[counting] % i == 0 && all_num[counting + 1] % i == 0) {
flag++;
}
if (flag == 1) {
return i;
}
}
return 1;
}
}
import java.util.Scanner;
public class Main {
//(1)求最大公约数法:最大公约数*最小公倍数 = 两个数之积。
//(2)使用辗转相除法求 比如:下图为求 2 4 6的最小公倍数,用2 4 6做辗转相除法可得最小公倍数为2*1*2
// 3.而第三种方法为:先求出n个数中最大的那个数max,如果这个数可以整除所有的数,则这是最小公倍数
// 如果不能则令max+1直到找到可以整除所有的数的那个数为止
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
while (sc_01.hasNext()) {
int char_num = sc_01.nextInt();
int[] all_num = new int[char_num];
for (int i = 0; i < char_num; i++) {
all_num[i] = sc_01.nextInt();
}
for (int i = 1; i < char_num; i++) {
if(all_num[i]>all_num[0]){
int max = all_num[0];
all_num[0] = all_num[i];
all_num[i] = max;
}
}
int least_multiple = all_num[0];
least_multiple = check_Mul(all_num,least_multiple);
System.out.println(least_multiple);
}
}
private static int check_Mul(int[] all_num, int least_multiple) {
while (true){
int couting =0;
for(int i = 0;i<all_num.length;i++){
if(least_multiple % all_num[i] == 0){
couting ++;
}
}
if(couting == all_num.length){
return least_multiple;
}else{
least_multiple++;
}
}
}
}
2029题:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
int char_num = sc_01.nextInt();
for(int i = 0;i<char_num;i++){
String test_strings = sc_01.next();
boolean result = check_huiwen(test_strings);
if(result){System.out.println("yes");}
else{
System.out.println("no");
}
}
}
private static boolean check_huiwen(String test_strings) {
for(int i= 0;i<test_strings.length();i++){
if(test_strings.charAt(i) != test_strings.charAt(test_strings.length()-1-i)){
return false;
}
}
return true;
}
}
2030题:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
int line_num = sc_01.nextInt();
int couting = 0;
sc_01.nextLine();
for (int i = 0; i < line_num; i++) {
String s1 = sc_01.nextLine();//正常输入流就可以
// 是通过Unicode进行汉字编码的,中文的unicode编码范围是: 在33-126之间的字母。
//只要>32和<126之内即可。
//链接:https://www.qqxiuzi.cn/zh/hanzi-unicode-bianma.php
for (int j = 0; j < s1.length(); j++) {
if ((int) s1.charAt(j) <32 ||((int) s1.charAt(j) >126)){
// if (((int) s1.charAt(j) >= 0x4E00 && (int) s1.charAt(j) <= 0x9FEF) ||
// ((int) s1.charAt(j) >= 0x3400 && (int) s1.charAt(j) <= 0x4DB5) ||
// ((int) s1.charAt(j) >= 0x20000 && (int) s1.charAt(j) <= 0x2A6D6) ||
// ((int) s1.charAt(j) >= 0x2A700 && (int) s1.charAt(j) <= 0x2B734) ||
// ((int) s1.charAt(j) >= 0x2B740 && (int) s1.charAt(j) <= 0x2B81D) ||
// ((int) s1.charAt(j) >= 0x2B820 && (int) s1.charAt(j) <= 0x2CEA1) ||
// ((int) s1.charAt(j) >= 0x2CEB0 && (int) s1.charAt(j) <= 0x2EBE0) ||
// ((int) s1.charAt(j) >= 0x2F00 && (int) s1.charAt(j) <= 0x2FD5) ||
// ((int) s1.charAt(j) >= 0x2E80 && (int) s1.charAt(j) <= 0x2EF3) ||
// ((int) s1.charAt(j) >= 0xF900 && (int) s1.charAt(j) <= 0xFAD9) ||
// ((int) s1.charAt(j) >= 0x2F800 && (int) s1.charAt(j) <= 0x2FA1D) ||
// ((int) s1.charAt(j) >= 0xE815 && (int) s1.charAt(j) <= 0xE86F) ||
// ((int) s1.charAt(j) >= 0xE400 && (int) s1.charAt(j) <= 0xE5E8) ||
// ((int) s1.charAt(j) >= 0xE600 && (int) s1.charAt(j) <= 0xE6CF) ||
// ((int) s1.charAt(j) >= 0x31C0 && (int) s1.charAt(j) <= 0x31E3) ||
// ((int) s1.charAt(j) >= 0x2FF0 && (int) s1.charAt(j) <= 0x2FFB) ||
// ((int) s1.charAt(j) >= 0x3105 && (int) s1.charAt(j) <= 0x312F) ||
// ((int) s1.charAt(j) >= 0x31A0 && (int) s1.charAt(j) <= 0x31BA) ||
// ((int) s1.charAt(j) == 0x3007)) {
couting++;
}
}
System.out.println(couting);
couting = 0;
}
}
}
2032题:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
while (sc_01.hasNext()) {
int line_nunm = sc_01.nextInt();
if (line_nunm == 1) {
System.out.println("1");
System.out.println();
} else if (line_nunm == 2) {
System.out.println("1");
System.out.println("1 1");
System.out.println();
} else {
System.out.println("1");
System.out.println("1 1");
int[] Inital_val = {1, 1};
int counting = 2;
while (counting < line_nunm) {
int[] Out_val = new int[Inital_val.length + 1];
Out_val[0] = 1;
Out_val[Out_val.length - 1] = 1;
//计算第i行的值。
for (int i = 1; i < Out_val.length - 1; i++) {
Out_val[i] = Inital_val[i] + Inital_val[i - 1];
}
//输出第i行结果
for (int i = 0; i < Out_val.length - 1; i++) {
System.out.print(Out_val[i] + " ");
}
System.out.println(Out_val[Out_val.length - 1]);
//更新旧Value的值
Inital_val = Out_val;
counting++;
}
System.out.println();
}
}
}
}
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
while (sc_01.hasNext()) {
int line_nunm = sc_01.nextInt();
int[][] a = new int[32][32];
for (int i = 1; i <= 30; i++) a[i][0] = 1;
for (int i = 2; i <= 30; i++)
for (int j = 1; j < i; j++) a[i][j] = a[i - 1][j - 1] + a[i - 1][j];
for (int i = 1; i <= line_nunm; i++) {
for (int j = 0; j < i - 1; j++)
System.out.print(a[i][j] + " ");
System.out.println(a[i][i - 1]);
}
System.out.println();
}
}
}
2040题:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
int line_num = sc_01.nextInt();
for(int i= 0;i<line_num;i++){
int num_01 = sc_01.nextInt();
int num_02 = sc_01.nextInt();
int result_01 = 0,result_02 = 0;
//计算其第一个数的真约数之和
for(int j=1;j<num_01;j++){
if(num_01%j==0){
result_01+=j;
}
}
//计算第二个数的真约数之和
for(int j=1;j<num_02;j++){
if(num_02%j==0){
result_02+=j;
}
}
if(num_01==result_02 && num_02 ==result_01){
System.out.println("YES");
}else {
System.out.println("NO");
}
}
}
}
2042题:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
int line = sc_01.nextInt();
for(int i =0;i<line;i++){
int toll_station = sc_01.nextInt();
int total_goat = 3;
for(int j = 0;j<toll_station;j++){
total_goat = (total_goat - 1) *2;
}
System.out.println(total_goat);
}
}
}
2054题:
import java.math.BigDecimal;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
while (sc_01.hasNext()) {
BigDecimal num_01 = sc_01.nextBigDecimal();// BigDecimal 由任意精度的整数非标度值 和 32 位的整数标度 (scale) 组成
BigDecimal num_02 = sc_01.nextBigDecimal();
if (num_01.compareTo(num_02) == 0) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
}
}
2055题:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc_01 = new Scanner(System.in);
int line_num = sc_01.nextInt();
for (int i = 0; i < line_num; i++) {
String string_01 = sc_01.next();
int y = sc_01.nextInt();
int result = 0;
if (string_01.charAt(0) >= 'a' && string_01.charAt(0) <= 'z') {
result = -(string_01.charAt(0) - 96);
} else if (string_01.charAt(0) >= 'A' && string_01.charAt(0) <= 'Z') {
result = (string_01.charAt(0) - 64);
} else {
}
result = result + y;
System.out.println(result);
}
}
}