密码要求:
1.长度超过8位
2.包括：大写字母/小写字母/数字/其它符号，以上四种至少三种
3.不能分割出两个相等的长度大于 2 的子串，例如 abcabc 可以分割出两个 abc，不合法，ababa 则无法分割出2个aba。
注：其他符号不含空格或换行
数据范围：输入的字符串长度满足 
[编程题]密码验证合格程序
import sys
# 判断密码是否合法
def is_legal(str1):
lst1 = list(str1)
# pop掉最后一个换行符
lst1.pop()
# 长度判断
if len(lst1) <= 8:
return 'NG'
# 公共子串判断
for i in range(len(lst1)-2):
# 长度为3的子串, 重复出现则返回NG
if str1[i:i+3] in str1[i+1:]:
return 'NG'
# 四种情况判断
kind_dict = {'dig': 0, 'upper': 0, 'lower': 0, 'other': 0}
for i in lst1:
# 数字
if i.isdigit():
kind_dict['dig'] = 1
# 大写字母
elif ord('Z') >= ord(i) and ord('A') <= ord(i):
kind_dict['upper'] = 1
# 小写字母
elif ord('z') >= ord(i) and ord('a') <= ord(i):
kind_dict['lower'] = 1
# 其他
else:
kind_dict['other'] = 1
# 满足三种以上
if sum(list(kind_dict.values())) >= 3:
return 'OK'
else:
return 'NG'
# 读取
lines = list(sys.stdin.readlines())
for i in lines:
print(is_legal(i))
发表于 2022-07-25 20:33:55
回复(0)
抄大佬的，第二个分割确实很巧妙，大佬们挺强的
def check_(s):
alpha=0
num=0
ALPHA=0
other=0
s=s.split()
s=s[0]
if len(s)<=8:
print('NG')
return None
s2=s[:]
c1=0
c2=1
for i in s:
if 'a'<=i<='z':
alpha=1
s=s[1:]
if '0'<=i<='9':
num=1
s=s[1:]
if 'A'<=i<='Z':
ALPHA=1
s=s[1:]
if len(s)>0:
other=1
if alpha+num+ALPHA+other>=3:
c1=1
for i in range(3,len(s2)):
if len(s2.split(s2[i-3:i]))>2:
c2=0
break
if c1+c2==2:
print('OK')
else:
print('NG')
import sys
for line in sys.stdin:
line=line[:-1]
check_(line)
发表于 2022-04-15 21:35:09
回复(0)
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
/**
* @author Yuliang.Lee
* @version 1.0
* @date 2021/8/29 9:46
* 校验密码有效性：
密码要求：
1.长度超过8位
2.包括大小写字母.数字.其它符号,以上四种至少三种
3.不能有相同长度大于2的子串重复
示例1
输入：
021Abc9000
021Abc9Abc1
021ABC9000
021$bc9000
输出：
OK
NG
NG
OK
*/
public class Main {
public static String REGEX_1 = "[A-Z]";
public static String REGEX_2 = "[a-z]";
public static String REGEX_3 = "\\d";
public static String REGEX_4 = "[^A-Za-z0-9]";
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
String password = in.nextLine();
if (password.length() <= 8) {
System.out.println("NG");
continue;
}
int countRegex = 0;
if (hasRegex(REGEX_1, password)) {
countRegex++;
}
if (hasRegex(REGEX_2, password)) {
countRegex++;
}
if (hasRegex(REGEX_3, password)) {
countRegex++;
}
if (hasRegex(REGEX_4, password)) {
countRegex++;
}
if (countRegex < 3) {
System.out.println("NG");
continue;
}
if (isSubRepeat(password)) {
System.out.println("NG");
continue;
}
System.out.println("OK");
}
}
public static boolean hasRegex(String regex, String input) {
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(input);
if (matcher.find()) {
return true;
}
return false;
}
/**
* 检查是否存在长度为3的多个相同的子串
* @param input
* @return
*/
public static boolean isSubRepeat(String input) {
for (int i = 0; i < input.length() - 3; i++) {
// 如果用模式regex相关的话还需要处理转译字符
if (input.substring(i + 1).contains(input.substring(i, i + 3))) {
return true;
}
}
return false;
}
}
发表于 2021-08-29 12:37:31
回复(0)
class Solution:
def func(self,m:str):
dic={}
a=[0 for _ in range(4)]
b=[0 for _ in range(len(m)-2)]
for eachstr in m:
if not eachstr.isalnum():
a[0]=1
elif eachstr.isdigit():
a[1]=1
elif eachstr.islower():
a[2]=1
elif eachstr.isupper():
a[3]=1
for i in range(len(b)):
b[i]=m[i:i+3]
if len(b) != len(set(b)):
return'NG'
if len(m)<=8:
return 'NG'
if a[0]+a[1]+a[2]+a[3]<=2:
return 'NG'
return 'OK'
while True:
try:
m=input()
print(Solution().func(m))
except:
break
def func(self,m:str):
dic={}
a=[0 for _ in range(4)]
b=[0 for _ in range(len(m)-2)]
for eachstr in m:
if not eachstr.isalnum():
a[0]=1
elif eachstr.isdigit():
a[1]=1
elif eachstr.islower():
a[2]=1
elif eachstr.isupper():
a[3]=1
for i in range(len(b)):
b[i]=m[i:i+3]
if len(b) != len(set(b)):
return'NG'
if len(m)<=8:
return 'NG'
if a[0]+a[1]+a[2]+a[3]<=2:
return 'NG'
return 'OK'
while True:
try:
m=input()
print(Solution().func(m))
except:
break
发表于 2021-08-27 17:19:20
回复(0)
import java.util.*;
public class Main{
public static Boolean checkPassword(String str){
// 验证长度
if(str.length() <= 8) return false;
// 验证包括的字符种类
int low = 0, up = 0, num = 0, other = 0;
for(int i = 0; i < str.length(); i++){
char tmp = str.charAt(i);
if(tmp >= 'a' && tmp <= 'z'){
low = 1;
}
else if(tmp >= 'A' && tmp <= 'Z'){
up = 1;
}
else if(tmp >= '0' && tmp <= '9'){
num = 1;
}
else{
other = 1;
}
}
if(low+up+num+other < 3) return false;
// 验证是否有长度大于2的子串重复
for(int i = 0; i < str.length(); i++){
for(int j = i+3; j < str.length(); j++){
String tmp = str.substring(i, j);
if(str.substring(j).contains(tmp)){
return false;
}
}
}
// 以上false都没有出现，返回true
return true;
}
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
String str = sc.next();
if(checkPassword(str)){
System.out.println("OK");
}
else{
System.out.println("NG");
}
}
}
}
发表于 2021-07-15 15:37:11
回复(0)
let line
const getTypes = password => {
let count = 0;
/[0-9]/.test(password) && count++;
/[a-z]/.test(password) && count++;
/[A-Z]/.test(password) && count++;
/[^A-Za-z0-9]/.test(password) && count++;
return count;
}
const hasRepeat = password => {
let arr = password.split('')
for(let i = 0; i < password.length - 2; i++) {
let subStr = arr.splice(i, 3, ' ').join('')
if(arr.join('').includes(subStr)) return false
arr = password.split('')
}
return true
}
while(line = readline()) {
let length = line.length > 8
let type = getTypes(line) >= 3
let repeat = hasRepeat(line)
let ans = (length && type && repeat) ? 'OK' : 'NG'
print(ans)
}
编辑于 2021-07-08 11:54:24
回复(0)
import java.util.Scanner;
import java.util.HashSet;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
String str = sc.nextLine();
boolean reslut = true;
HashSet<String> hm = new HashSet();
if(str.length() <= 8){ //长度小于8
reslut = false;
}else{//判断包含大小写，数字，特殊符号
// -1为保留尾部空元素
if (str.split("\\d",-1).length >= 2) hm.add("number");
if (str.split("[a-z]",-1).length >= 2) hm.add("lower");
if (str.split("[A-Z]",-1).length >= 2) hm.add("Upper");
if (str.split("[\\W_]",-1).length >= 2) hm.add("ha?");
}
if(hm.size() < 3){//少于三种
reslut = false;
}
//判断是否有重复字符串
for (int i = 0; i <= str.length()-3; i++){
String string = str.substring(i,i+3);
if(str.substring(i+1).contains(string)){
reslut = false;
}
}
if(reslut){
System.out.println("OK");
}else{
System.out.println("NG");
}
}
}
}
发表于 2021-06-12 17:12:16
回复(0)
import sys
import re
def func(s):
if len(s) <=8 :
return 'NG'
arr = [0, 0, 0, 0]
for c in s:
if c.isupper():
arr[0] = 1
elif c.islower():
arr[1] = 1
elif c.isdigit():
arr[2] = 1
else:
arr[3] = 1
if sum(arr) < 3:
return 'NG'
if re.findall(r'(.{3,}).*\1', s):
return 'NG'
return 'OK'
for line in sys.stdin:
print(func(line.strip()))
发表于 2021-05-04 11:20:28
回复(0)
条件三：只需要找有无长度为3的字串重复
public static void main(String[] args){
Scanner in = new Scanner(System.in);
while(in.hasNext()){
String str = in.nextLine();
//条件一
if(str.length() <= 8){
System.out.println("NG");
continue;
}
boolean[] flag = new boolean[4];//判断条件二
HashSet<String> set = new HashSet<>();//判断条件三
boolean is = true;//记录for循环里是否有输出
for(int i = 0;i < str.length();i++){
char c = str.charAt(i);
if(i + 2 < str.length()){
String cur = str.substring(i,i + 3);
if(set.contains(cur)){
System.out.println("NG");
is = false;
break;
}else{
set.add(cur);
}
}
if(c >= '0' && c <= '9'){
flag[0] = true;
}else if(c >= 'a' && c <= 'z'){
flag[1] = true;
}else if(c >= 'A' && c <= 'Z'){
flag[2] = true;
}else{
flag[3] = true;
}
}
if(!is){//for循环里有输出
continue;
}
int count = 0;
for(int i = 0;i < flag.length;i++){
if(flag[i] == false){
count++;
}
}
if(count > 1){
System.out.println("NG");
continue;
}
System.out.println("OK");
}
}
发表于 2021-04-18 10:22:10
回复(0)
依题意设立三道检查逻辑即可。麻烦一些的点是检查某个子串是否重复，可以从长度为3的子串开始，利用hash表，在原密码字符串中滑窗检查其是否重复。这里可以对检查的逻辑进行剪枝，超过原来字符串长度一半的子串就不用检查了，这样的子串是不可能重复的。
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.HashSet;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String passWord;
while((passWord = br.readLine()) != null){
if(judge(passWord))
System.out.println("OK");
else
System.out.println("NG");
}
}
private static boolean judge(String pwd) {
// 长度不足8
if(pwd.length() <= 8) return false;
int lowerAlphaCnt = 0, upperAlphaCnt = 0, digit = 0, other = 0;
for(int i = 0; i < pwd.length(); i++){
if(pwd.charAt(i) >= 'A' && pwd.charAt(i) <= 'Z')
upperAlphaCnt ++;
else if(pwd.charAt(i) >= 'a' && pwd.charAt(i) <= 'z')
lowerAlphaCnt ++;
else if(pwd.charAt(i) >= '0' && pwd.charAt(i) <= '9')
digit ++;
else
other ++;
}
// 字符种类不足三种
if(Math.min(lowerAlphaCnt, 1) + Math.min(upperAlphaCnt, 1) + Math.min(digit, 1) + Math.min(other, 1) < 3)
return false;
// 看是否有相同长度大于2的子串重复
HashSet<String> set = new HashSet<>();
for(int len = 3; len < pwd.length() / 2; len++){
for(int i = 0; i <= pwd.length() - len; i++){
String substr = pwd.substring(i, i + len);
if(!set.contains(substr)){
set.add(substr);
}else{
// 发现已经有这个子串了
return false;
}
}
}
return true;
}
}
发表于 2021-03-22 14:57:05
回复(0)
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
// 注意 hasNext 和 hasNextLine 的区别
while (in.hasNextLine()) { // 注意 while 处理多个 case
String pwd = in.nextLine();
HashMap<Integer,Boolean> map = new HashMap<>(4);
//1.长度超过8位
if(pwd.length() <= 8){
System.out.println("NG");
continue;
}
//2.包括大小写字母.数字.其它符号,以上四种至少三种
for(int i = 0; i < pwd.length(); i++){
char c = pwd.charAt(i);
//有三种了
if(map.size() == 3) break;
//小写字母
if(c >= 'a' && c <= 'z'){
map.put(0,true);
}
//大写字母
else if(c >= 'A' && c <= 'Z'){
map.put(1,true);
}
//数字
else if(c >= '0' && c <= '9'){
map.put(2,true);
}
else{
map.put(3,true);
}
}
if(map.size() < 3) {
System.out.println("NG");
continue;
}
//3. 不能有相同长度大于2的子串重复。
//也就是说在这里，我们就只需要考虑长度为3的子字符串,是否有重复
boolean flag = false;
for(int j = 0; j <= pwd.length() - 6; j++){
String sub = pwd.substring(j , j+3);
if(pwd.indexOf(sub,j+3) != -1) {
flag = true;
break;
}
}
if(flag){
System.out.println("NG");
continue;
}
System.out.println("OK");
}
}
}
没什么难点，第三条规则是不能有相同长度大于2的子字符串，那么我们直接看长度为3的子字符串就可以了。个人认为这种长度的子字符串，就不用考虑什么KMP算法了，没太大区别的，当然这里我用的indexOf()本来就是KMP实现的😜。
另外一个容易纠结的点就是比如00000这种，是否算重复。本题是不算的，也就是说，两个重复的子字符串不可以重叠。
发表于 2021-01-02 19:20:05
回复(0)
妹的, 是我理解有误还是题出的不对
3.不能有相同长度大于等于2的子串重复
但是测试集实际的要求是"大于2"
上面测试样本有长度为2的重复, 给的结果还是 "OK"4@M$68(Oh%!n%~9&08&Z@#+dN0&Z
发表于 2020-07-02 14:44:24
回复(1)
int main()
{
string a;
while(cin>>a)
{
int flat[4]={0};
if(a.size()<=8)
{
cout<<"NG"<<endl;
continue;
}
else
{
int j=0;
for(int i=0;i<a.size();i++)
{
if(isdigit(a[i]))
flat[0]=1;
else if(islower(a[i]))
flat[1]=1;
else if(isupper(a[i]))
flat[2]=1;
else
flat[3]=1;
if(flat[0]+flat[1]+flat[2]+flat[3]==3)
break;
}
if(flat[0]+flat[1]+flat[2]+flat[3]==3)
{
int i=2;
for(;i<a.size()-2;i++)
{
if(a.find(string(a.begin()+i-2,a.begin()+i+1),i+1)!=string::npos)
{
cout<<"NG"<<endl;
break;
}
}
if(i==a.size()-2)
cout<<"OK"<<endl;
}
else
cout<<"NG"<<endl;
}
}
return 0;
} 先检测长度，然后检测字符是否三种以上，然后检测重复，都满足则欧克
发表于 2020-03-19 12:07:50
回复(0)
难度不大，思路也比较好想出来。
#include <stdio.h>
(737)#include <string.h>
#include <ctype.h>
int main()
{
int i,j,n;
char keywords[128]={0};
int num[4]={0};
sc: while(scanf("%s",keywords) != EOF)
{
n = strlen(keywords);
num[0] = num[1] = num[2] = num[3] = 0;
if( n <= 8)
{
printf("NG\n");
continue;
}
for(i=0; keywords[i]!=0; i++)
{
if(islower(keywords[i]) )
num[0]= 1;
else if(isupper(keywords[i]))
num[1]= 1;
else if(isdigit(keywords[i]))
num[2]= 1;
else if(isgraph(keywords[i]))
num[3]= 1;
}
if(num[0]+num[1]+num[2]+num[3] < 3)
{
printf("NG\n");
continue;
}
for(i=0; i+2 < n; i++)
for(j=1; j+2 < n;j++)
if(i!=j && keywords[i] == keywords[j] && keywords[i+1] == keywords[j+1] \
&& keywords[i+2] == keywords[j+2])
{
printf("NG\n");
goto sc;
}
printf("OK\n");
}
return 0;
}
发表于 2020-03-18 18:08:29
回复(0)
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
String x = sc.nextLine();
int[] flag = new int[4];
if (x.length() <= 8) {
System.out.println("NG");
continue;
}
for (int i = 0; i < x.length(); i++) {
if (x.charAt(i) >= 'a' && x.charAt(i) <= 'z') {
flag[0] = 1;
} else if (x.charAt(i) >= 'A' && x.charAt(i) <= 'Z') {
flag[1] = 1;
} else if (x.charAt(i) >= '0' && x.charAt(i) <= '9') {
flag[2] = 1;
} else {
flag[3] = 1;
}
}
if (flag[0] + flag[1] + flag[2] + flag[3] < 3) {
System.out.println("NG");
continue;
}
boolean valid = true;
for1:
for (int i = 0; i <= x.length() - 6; i++) {
for2:
for (int j = i + 3; j <= x.length() - 3; j++) {
if (x.substring(i, i + 3).equals(x.substring(j, j + 3))) {
System.out.println("NG");
valid = false;
break for1;
}
}
}
if (valid) {
System.out.println("OK");
}
}
}
}
发表于 2020-03-07 18:25:31
回复(1)
这道题比较难得就是字符子串超过2的判断吧，两个for循环 第一次循环判断+3对比，之后挨个加，如果有相等的就直接退出打个标签。
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<sstream>
#include<algorithm>
using namespace std;
bool beyond2(string s)
{
int len=s.length();
int flag=0;
for(int i=0; i<len-3; i++)
{
for(int j=i+3; j<=len-3; j++)
{
string temp1=s.substr(i,3);
string temp2=s.substr(j,3);
if(temp1==temp2)
{
flag=1;
// cout<<"temp1: "<<temp1<<" temp2: "<<temp2<<endl;
break;
}
}
if(flag==1)
break;
}
if(flag==0)
return true;
//cout<<"OK"<<endl;
else
return false;
//cout<<"NG"<<endl;
}
bool beyond8(string s)
{
int len=s.length();
if(len>8)
return true;
return false;
}
bool isAa(string s)
{
int len=s.length();
int sum=0;
int flag0=0,flag1=0,flag2=0,flag3=0,flag4=0;
for(int i=0; i<len; i++)
{
if(s[i]>='a'&&s[i]<='z')
flag0=1;
else if(s[i]>='A'&&s[i]<='Z')
flag1=1;
else if(s[i]>='0'&&s[i]<='9')
flag2=1;
else if(s[i]>='a'&&s[i]<='z')
flag3=1;
else
flag4=1;
}
if((flag0+flag1+flag2+flag3+flag4)>=3)
return true;
return false;
}
int main()
{
string s;
while(getline(cin,s))
{
if(beyond8(s)&&beyond2(s)&&isAa(s))
{
cout<<"OK"<<endl;
}
else
cout<<"NG"<<endl;
}
return 0;
}
发表于 2019-10-04 00:54:10
回复(1)
知识点:
1.cctype.h
isspace(),isdigit(),isalpha(),ispunct(),islower(),isupper(),tolower(),toupper();
2.string.h
getline(cin,str),getline(cin,str,ch); 字符串长度: int len=0,while(str[++len]);
解题思路:
1.读取字符串str,计算字符串长度len；
2.利用cctype.h头文件中的函数判断字符类型，定义一个数组类存储字符串中大写字母，小写字母，数字和其他符号的个数；
3.循环判断是否有长度为3的子串重复，先判断子串str1=(str[0],str[1],str[2])和str2=(str[3],str[4],str[5])，然后判断子串str1=(str[1],str[2],str[3])和str2=(str[4],str[5],str[6]),依次循环判断直到字符串尾。(注意str1子串循环范围i:0-len-6,str2子串循环范围j=i+3:len)
4.若len>8,步骤2数组非零取值大于等于3，步骤3条件不出现子串重复，则输出“OK”，否则输出“NG”
C++代码实现:
#include<iostream>
#include<cctype>
#include<string>
using namespace std;
bool code(string str);
bool Substring(string str,int n);
int main()
{
string str;
while(getline(cin,str))
{
if(code(str))
{
cout<<"OK"<<endl;
}
else
{
cout<<"NG"<<endl;
}
}
return 0;
}
bool code(string str)
{
int len=0,i;
while(str[++len]);
int temp[4]={0,0,0,0};
for(i=0;i<len;i++)
{
if(isdigit(str[i])) temp[0]++;
else if(islower(str[i])) temp[1]++;
else if(isupper(str[i])) temp[2]++;
else temp[3]++;
}
int K=0;
for(i=0;i<4;i++)
{
if(temp[i]>0)
K++;
}
if((K>=3) && (len>8) && (Substring(str,len)))
return true;
else
return false;
}
bool Substring(string str, int n)
{
bool flag=true;
for(int i=0;i<= n-6;i++)
{
for(int j=i+3;j<n;j++)
{
if(str[i]==str[j] && str[i+1]==str[j+1] && str[i+2]==str[j+2])
{
flag= false;
break;
}
}
}
return flag;
}
#include<cctype>
#include<string>
using namespace std;
bool code(string str);
bool Substring(string str,int n);
int main()
{
string str;
while(getline(cin,str))
{
if(code(str))
{
cout<<"OK"<<endl;
}
else
{
cout<<"NG"<<endl;
}
}
return 0;
}
bool code(string str)
{
int len=0,i;
while(str[++len]);
int temp[4]={0,0,0,0};
for(i=0;i<len;i++)
{
if(isdigit(str[i])) temp[0]++;
else if(islower(str[i])) temp[1]++;
else if(isupper(str[i])) temp[2]++;
else temp[3]++;
}
int K=0;
for(i=0;i<4;i++)
{
if(temp[i]>0)
K++;
}
if((K>=3) && (len>8) && (Substring(str,len)))
return true;
else
return false;
}
bool Substring(string str, int n)
{
bool flag=true;
for(int i=0;i<= n-6;i++)
{
for(int j=i+3;j<n;j++)
{
if(str[i]==str[j] && str[i+1]==str[j+1] && str[i+2]==str[j+2])
{
flag= false;
break;
}
}
}
return flag;
}
发表于 2019-08-30 09:35:34
回复(0)
import java.util.Scanner;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.regex.Pattern;
public class Main {
public static boolean checkLength(String password) { if (password.length() > 8) return true; else return false; } public static boolean checkCharKinds(String password) { int Digit = 0, lowercase = 0, uppercase = 0, others = 0; Pattern pattern1 = Pattern.compile("[0-9]*");// 正则表达式 Pattern pattern2 = Pattern.compile("[a-z]*");// 正则表达式 Pattern pattern3 = Pattern.compile("[A-Z]*");// 正则表达式 for (int i = 0; i < password.length(); i++) { if (pattern1.matcher(String.valueOf(password.charAt(i))).matches()) Digit = 1; else if (pattern2.matcher(String.valueOf(password.charAt(i))).matches()) lowercase = 1; else if (pattern3.matcher(String.valueOf(password.charAt(i))).matches()) uppercase = 1; else others = 1; } if ((Digit + lowercase + uppercase + others) >= 3) return true; else return false; } public static boolean checkCharRepeat(String password) { for (int i = 0; i < password.length() - 2; i++) { String substr1 = password.substring(i, i + 3); if (password.substring(i + 1).contains(substr1)) return false; } return true; } public static void main(String[] args) { Scanner cin = new Scanner(System.in); while (cin.hasNextLine()) { String psw = cin.nextLine(); if (checkLength(psw) && checkCharKinds(psw) && checkCharRepeat(psw)) System.out.println("OK"); else System.out.println("NG"); } }
}
发表于 2019-06-22 20:41:38
回复(0)
#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main(void)
{
string password;
pair<string, string> dout("OK", "NG");
vector<string> out;
while (cin >> password) {
if (password.size() <= 8) {//长度小于等于8
out.push_back(dout.second);
}
else {
char count[4] = { 0 };
for (int i = 0; i < password.size(); ++i) {
if (password[i] >= '0' && password[i] <= '9') {
count[0] = 1;
}
else if (password[i] >= 'a' && password[i] <= 'z') {
count[1] = 1;
}
else if (password[i] >= 'A' && password[i] <= 'Z') {
count[2] = 1;
}
else {
count[3] = 1;
}
}
if ((count[0] + count[1] + count[2] + count[3]) < 3) {
out.push_back(dout.second); //密码种类必须三种以上
}
else {
bool flag = 0;
for (int i = 0; i < password.size() - 6; ++i) {
for (int j = i + 3; j < password.size() - 2; ++j) {
if ((password[i] == password[j]) && (password[i + 1] == password[j + 1]) && (password[i + 2] == password[j + 2])) {
flag = 1;
}
}
}
if (flag == 0) {
out.push_back(dout.first);//没问题
}
else {
out.push_back(dout.second);//出现重复子串
}
}
}
}
for (auto i = out.begin(); i < out.end(); ++i) {
cout << *i << endl;
}
return 0;
}
#include <vector>
#include <string>
using namespace std;
int main(void)
{
string password;
pair<string, string> dout("OK", "NG");
vector<string> out;
while (cin >> password) {
if (password.size() <= 8) {//长度小于等于8
out.push_back(dout.second);
}
else {
char count[4] = { 0 };
for (int i = 0; i < password.size(); ++i) {
if (password[i] >= '0' && password[i] <= '9') {
count[0] = 1;
}
else if (password[i] >= 'a' && password[i] <= 'z') {
count[1] = 1;
}
else if (password[i] >= 'A' && password[i] <= 'Z') {
count[2] = 1;
}
else {
count[3] = 1;
}
}
if ((count[0] + count[1] + count[2] + count[3]) < 3) {
out.push_back(dout.second); //密码种类必须三种以上
}
else {
bool flag = 0;
for (int i = 0; i < password.size() - 6; ++i) {
for (int j = i + 3; j < password.size() - 2; ++j) {
if ((password[i] == password[j]) && (password[i + 1] == password[j + 1]) && (password[i + 2] == password[j + 2])) {
flag = 1;
}
}
}
if (flag == 0) {
out.push_back(dout.first);//没问题
}
else {
out.push_back(dout.second);//出现重复子串
}
}
}
}
for (auto i = out.begin(); i < out.end(); ++i) {
cout << *i << endl;
}
return 0;
}
发表于 2019-04-14 18:09:17
回复(0)
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