题目2
题目4
1.恺撒密码
输入：
偏移量K
待加密的字符串
输出：
加密后的字符串。注意只需用对大小写字母进行变换，其它字符原样输出
public class Main{
public static void main(String[] args) {
// 偏移大小
int offset = Integer.parseInt(StdIn.readLine());
while (!StdIn.isEmpty()) {
String inStr = StdIn.readLine();
char[] arr = inStr.toCharArray();
// 对单个字符解密
for (int i = 0; i < arr.length; i++) {
int chNum = arr[i];
int passChar = 0;
// 是否为英文字母
if (chNum >= 97 && chNum <= 122) {
passChar = chNum + offset;
if (passChar > 122) {
passChar = passChar - 123 + 97;
}
} else if (chNum >= 65 && chNum <= 90) {
passChar = chNum + offset;
if (passChar > 90) {
passChar = passChar - 91 + 65;
}
} else {
passChar = chNum;
}
arr[i] = (char) passChar;
}
// 输出
System.out.println(arr);
}
}
}
2.翻硬币
输入：
初始硬币的数量n
输出：
至少需要多少次翻转，又能恢复最初的正面朝上。
public class Main {
public static void main(String[] args) {
// 输入硬币数量
int N = StdIn.readInt();
// 定义硬币数字，false:正面 true:背面
boolean[] bool = new boolean[N];
// 统计反转次数
int num = 0;
// 循环控制条件
boolean breakFlag = true;
while (breakFlag) {
for (int i = 0; i < N && breakFlag; i++) {
breakFlag = false;
int j = 0;
for (; j <= i / 2; j++) {
// 确定交换位置并交换
int x = i - j;
if (x == j) {
bool[j] = !bool[j];
} else {
boolean tmp = !bool[j];
bool[j] = !bool[x];
bool[x] = tmp;
}
}
num++;
// 检测退出条件成立
for (int k = 0; k < N; k++) {
if (bool[k]) {
breakFlag = true;
break;
}
}
}
}
System.out.println(num);
}
}
黄一鹤提供版本:
import java.util.Scanner;
public class Coins {
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
System.out.println("请输入硬币个数");
int coins=cin.nextInt();
int[] ints =new int[coins] ;
for (int i = 0; i < ints.length; i++) {
ints[i]=0;
}
boolean isTrue = false;
int i = 1;
for (; ; i++) {
isTrue = true;
int n = i%coins;
if(n==0) {
n=coins;
}
changePosition(ints,n);
trunCoins(ints,n);
printResult(ints,i);
for (int j = 0; j< ints.length; j++) {
if(ints[j]!=0) {
isTrue = false;
break;
}
}
if(isTrue) {
break;
}
}
System.out.println("```````````````````````````");
System.out.println("总共"+coins+"个硬币");
System.out.println("总共需要翻转"+i+"次");
System.out.println("```````````````````````````");
}
public static void changePosition(int[] ints,int position) {
position = position -1;;
int i = 0;
int temp;
while(i<position) {
temp = ints[i];
ints[i]=ints[position];
ints[position]=temp;
i++;
position--;
}
}
public static void trunCoins(int[] ints, int n) {
for (int j = 0; j < n; j++) {
ints[j] = (ints[j]+1)%2;
}
}
public static void printResult(int[] ints,int n) {
System.out.print("第"+n+"次翻转结果：");
for (int j = 0; j< ints.length; j++) {
System.out.print(" "+ints[j]);
}
System.out.println();
}
}
3.随机播放和生日“悖论”
输入：
歌曲总数量N，蒙特卡洛模拟的次数T（都是整数）
输出：
取整后的平均次数（为了方便系统进行自动判定，故请输出整数。取整方法：(int)(x+0.5)）
public class Main {
public static void main(String[] args) {
int N = StdIn.readInt();
int M = StdIn.readInt();
boolean arr[] = null;
double sum = 0;
for (int i = 0; i < M; i++) {
int num = 1;
arr = new boolean[N];
while (true) {
int rad = (int) (Math.random() * N);
if(arr[rad]){
break;
}
arr[rad] = true;
num ++;
}
sum += num;
}
System.out.println((int) (sum / M + 0.5));
}
}
黄一鹤提供版本：
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class MathRandom {
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
int N=cin.nextInt();
int T=cin.nextInt();
double sumTimes = 0;
int result;
for (int i = 0; i < T; i++) {
result = returnMandomTimes(N);
sumTimes = sumTimes + result;
}
System.out.println((int)(sumTimes/T+0.5));
}
public static int returnMandomTimes(int n) {
Map<Integer, String> map = new HashMap<Integer, String>();
boolean isExist = false;
int i = 1;
while (!isExist) {
int mathRandom = (int) (Math.random()*n);
if(null!=map.get(mathRandom)) {
isExist = true;
}
else {
map.put(mathRandom, "exist");
i++;
}
}
return i;
}
}