文档内容
闭卷编程考试
第一题：读取输入字符串存储在数组中，将每个字符串按照8个字符切割，如果不足8个
字符则自动补全为0，存储在新的字符串数组中，最后字符串按照升序输出。
测试样例：
输入：2 123456789 abc
输出：12345678 90000000 abc00000
（通过率只有80%，不知道为啥）
import java.util.*;
public class Main {
static List list = new ArrayList<>();
public static boolean CompareString(String s1,String s2){
char[] m1 = s1.toCharArray();
char[] m2 = s2.toCharArray();
for(int i = 0;i<8;i++){
if(m1[i] >= m2[i]){
return true;
}
else{//小于
return false;
}
}
return false;
}
public static String addzero(String result,int len){
for(int i=0;i8){
//继续切割
int l = result.length();
list.add(result.substring(0,8));
result = result.substring(8,l);
}
if(result.length()>0){
list.add(addzero(result,8-result.length()));
}
}
public static void main(String[] args) {//如何实现升序？
Scanner in = new Scanner(System.in);
int num = Integer.valueOf(in.next());
String[] s = new String[num];
// String[] sresult = new String[num*13];
for(int i=0;i8){//按照8个8个切割，直到没有8个
splitby8(s[i]);
}
if(sl<8){
//自动补全0
list.add(addzero(s[i],8-sl));
}
}
String[] sresult = new String[list.size()];
for(int i =0;i=0;j--){
ch2[i++] = ch1[j];
}
return ch2;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String sum = in.next();
char[] my = sum.toCharArray();
String result = "";
//得到正序，括号里面可以出现括号，肯定先处理里面括号
int i =0;
//如果有大括号
if(sum.contains("{")){
while(i='0'&& my[i]<='9'){//遇到number需要处理
String mynum = "";
int mk = i;
while(my[mk] != '{'){
mynum += my[mk++];
}
int num = Integer.valueOf(mynum);
int h = mk+1;
String temp = "";
while(my[h] != '}'){
temp += my[h];
h++;
}
for(int k=0;k='a'&&my[i]<= 'z'){//a-z
result += my[i++];
}
else if(my[i]>= 'A'&&my[i]<='Z'){//A-Z
result += my[i++];
}
else if(my[i] != '{'&&my[i] != '}'){
result += my[i++];
}
else{
i++;
}
}
}
//如果有中括号
i = 0;
if(result.isEmpty()) {
result = sum;
}
if(sum.contains("[")) {
char[] my1 = result.toCharArray();
String result1 = "";
while (i < my1.length) {
if (my1[i] >= '0' && my1[i] <= '9') {//遇到number需要处理
String my1num = "";
int mk = i;
while(my1[mk] != '['){
my1num += my1[mk++];
}
int num = Integer.valueOf(my1num);
int h = mk+1;
String temp = "";
while (my1[h] != ']') {
temp += my1[h];
h++;
}
for (int k = 0; k = 'a' && my1[i] <= 'z') {//a-z
result1 += my1[i++];
} else if (my1[i] >= 'A' && my1[i] <= 'Z') {//A-Z
result1 += my1[i++];} else if (my1[i] != '[' && my1[i] != ']') {
result1 += my1[i++];
}
else{
i++;
}
}
result = result1;
}
i = 0;
if(result.isEmpty()) {
result = sum;
}
char[] my2 = result.toCharArray();
String result2 = "";
//后处理小括号
//如何找到括号最近的数字
while(i '0' && my2[i] <= '9') {//遇到number需要处理
String my2num = "";
int mk = i;
while (my2[mk] != '(') {
my2num += my2[mk++];
}
int num = Integer.valueOf(my2num);
int h = mk+1;
String temp = "";
while (my2[h] != ')') {
temp += my2[h];
h++;
}
for (int k = 0; k < num; k++) {
result2 += temp;
}
i = h + 1;
}
else if(my2[i]>='a'&&my2[i]<= 'z'){//a-z
result2 += my2[i];
}
else if(my2[i]>= 'A'&&my2[i]<='Z'){//A-Zresult2 += my2[i];
}
else if(my2[i] != ')'&&my2[i] != '('){
result2 += my2[i];
}
i++;
}
//逆序输出
char[] ch2 = reverseChar(result2.toCharArray());
String s = new String(ch2);
System.out.println(s);
}
}
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第三题：破解小王的加密算法（没有时间做了……）
