CMB5. 卡中心密码安全规范
描述
小明在卡中心工作，用到的很多系统账号都需要设置安全密码。密码如果符合以下规范可以称为安全密码：输入描述
输入为一个字符串作为密码。输出描述
输出为将该密码改为安全密码的最小改变次数。如果它已经是安全密码，则返回0。示例1
输入:
aB3ab3
输出:
0
示例2
输入:
aaaaaa
输出:
2
示例3
输入:
abcdefG
输出:
1
C 解法, 执行用时: 2ms, 内存消耗: 364KB, 提交时间: 2019-03-30
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int main()
{
char C[100], S[21];
gets(C);
int low_c = 0, up_c = 0, num = 0, flag = 1;
int i, n, count = 0, n1 = 0, j = 0, n2 = 0;
if(strlen(C) > 20)
{
for(i = 0; i < strlen(C); i++)
{
if(C[i] == C[i + 1] && C[i + 1] == C[i + 2])
{
if(strlen(C) - n1 == 20)
{
S[j] = C[i];
j ++;
}
else
{
n1 ++;
S[j] = C[i];
S[j + 1] = C[i + 1];
j = j + 2;
i = i + 2;
}
}
else
{
S[j] = C[i];
j ++;
}
}
//printf("%d\n", j);
}
else
strcpy(S, C);
if(strlen(C) - n1 > 20)
{
strncpy(S, C, 20);
n2 = strlen(C) - n1 - 21;
}
//printf("%d\n", n2);
for(i = 0; i < strlen(C) - n1; i++)
{
if(S[i] >= 'a' && S[i] <= 'z')
{
low_c = 1;
}
if(S[i] >= 'A' && S[i] <= 'Z')
{
up_c = 1;
}
if(S[i] >= '0' && S[i] <= '9')
{
num = 1;
}
if(S[i] == S[i + 1] && S[i + 1] == S[i + 2])
{
i = i + 2;
flag = 0;
count ++;
}
}
if(flag == 0)
{
printf("%d", count + n1 + n2);
return 0;
}
n = 3 - low_c - up_c - num;
printf("%d", n + n1 + n2);
}
C 解法, 执行用时: 2ms, 内存消耗: 376KB, 提交时间: 2020-12-23
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int main()
{
char C[100], S[21];
gets(C);
int low_c = 0, up_c = 0, num = 0, flag = 1;
int i, n, count = 0, n1 = 0, j = 0, n2 = 0;
if(strlen(C) > 20)
{
for(i = 0; i < strlen(C); i++)
{
if(C[i] == C[i + 1] && C[i + 1] == C[i + 2])
{
if(strlen(C) - n1 == 20)
{
S[j] = C[i];
j ++;
}
else
{
n1 ++;
S[j] = C[i];
S[j + 1] = C[i + 1];
j = j + 2;
i = i + 2;
}
}
else
{
S[j] = C[i];
j ++;
}
}
}
else
strcpy(S, C);
if(strlen(C) - n1 > 20)
{
strncpy(S, C, 20);
n2 = strlen(C) - n1 - 21;
}
for(i = 0; i < strlen(C) - n1; i++)
{
if(S[i] >= 'a' && S[i] <= 'z')
{
low_c = 1;
}
if(S[i] >= 'A' && S[i] <= 'Z')
{
up_c = 1;
}
if(S[i] >= '0' && S[i] <= '9')
{
num = 1;
}
if(S[i] == S[i + 1] && S[i + 1] == S[i + 2])
{
i = i + 2;
flag = 0;
count ++;
}
}
if(flag == 0)
{
printf("%d", count + n1 + n2);
return 0;
}
n = 3 - low_c - up_c - num;
printf("%d", n + n1 + n2);
}
C 解法, 执行用时: 2ms, 内存消耗: 376KB, 提交时间: 2019-08-19
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int main()
{
char C[100], S[21];
gets(C);
int low_c = 0, up_c = 0, num = 0, flag = 1;
int i, n, count = 0, n1 = 0, j = 0, n2 = 0;
if(strlen(C) > 20)
{
for(i = 0; i < strlen(C); i++)
{
if(C[i] == C[i + 1] && C[i + 1] == C[i + 2])
{
if(strlen(C) - n1 == 20)
{
S[j] = C[i];
j ++;
}
else
{
n1 ++;
S[j] = C[i];
S[j + 1] = C[i + 1];
j = j + 2;
i = i + 2;
}
}
else
{
S[j] = C[i];
j ++;
}
}
}
else
strcpy(S, C);
if(strlen(C) - n1 > 20)
{
strncpy(S, C, 20);
n2 = strlen(C) - n1 - 21;
}
for(i = 0; i < strlen(C) - n1; i++)
{
if(S[i] >= 'a' && S[i] <= 'z')
{
low_c = 1;
}
if(S[i] >= 'A' && S[i] <= 'Z')
{
up_c = 1;
}
if(S[i] >= '0' && S[i] <= '9')
{
num = 1;
}
if(S[i] == S[i + 1] && S[i + 1] == S[i + 2])
{
i = i + 2;
flag = 0;
count ++;
}
}
if(flag == 0)
{
printf("%d", count + n1 + n2);
return 0;
}
n = 3 - low_c - up_c - num;
printf("%d", n + n1 + n2);
}
C++14 解法, 执行用时: 2ms, 内存消耗: 380KB, 提交时间: 2020-08-11
#include <iostream>
#include <string>
using namespace std;
int main(){
string s;
cin >> s;
int len = s.size();
int lower = 0, upper = 0, num = 0;
int cnt = 0, rep = 0;
int res = 0;
for(int i = 0; i < s.size(); ++i){
if(cnt > 0 && s[i] == s[i-1]){
++cnt;
if(cnt == 3){
++res;
if(len > 20){//同时减少达到3个连续字符处的重复字符（最划算，同时解决1、3）
--len;
cnt = 2;
}
else{
cnt = 0;
++rep;//多少组达到3的连续字符
}
}
}
else cnt = 1;
if(islower(s[i])) lower = 1;
else if(isupper(s[i])) upper = 1;
else if(s[i] >= '0' && s[i] <= '9') num = 1;
}
int f =3-lower-upper-num;
if(rep >= f) f = 0;//需要以插入形式得到的之前没有的大写/小写/数字（不超过rep的部分可以通过替换得到，同时解决2、3）
else f -= rep;
if(len < 6){
res += std::max(6-len, f);
}
else if(len > 20){
res += len - 20;//删除
res += f;//插入
}
else res += f;
cout<<res<<endl;
return 0;
}
C++ 解法, 执行用时: 2ms, 内存消耗: 408KB, 提交时间: 2020-08-19
#include <iostream>
#include<bits/stdc++.h>
using namespace std;
int main()
{
int len,i,j,ans,res3,res2,res1,res4,res5,res6;
bool n,d,x;
char s[1000];
gets(s);
len=strlen(s);
n=d=x=0;
res3=res4=res5=res6=0;
j=-1;
for(i=0;i<len;i++)
{
if('A'<=s[i]&&s[i]<='Z'&&d==0)
{
d=1;
}
if('a'<=s[i]&&s[i]<='z'&&x==0)
{
x=1;
}
if('0'<=s[i]&&s[i]<='9'&&n==0)
{
n=1;
}
if(i>=2)
{
if(s[i]==s[i-1]&&s[i]==s[i-2]&&s[i]==s[i+1]&&s[i]==s[i+2]&&i<len-2)
{
if(i-j>=3)
{
res6++;
}
}
if(s[i]==s[i-1]&&s[i]==s[i-2]&&s[i]==s[i+1]&&i<len-1)
{
if(i-j>=3)
{
res5++;
}
}
if(s[i]==s[i-1]&&s[i]==s[i-2])
{
if(i-j>=3)
{
res3++;
j=i;
}
res4++;
}
}
}
res2=0;
res3=res3-res5;
res5=res5-res6;
if(d==0) res2++;
if(x==0) res2++;
if(n==0) res2++;
if(len<6)
{
res1=6-len;
ans=max(res1,res3+res5+res6);
ans=max(ans,res2);
}
else
{
if(len>20)
{
res1=len-20;
//printf("%d %d\n",res1,res4);
if(res1>=res4)
{
ans=res1+res2;
}
else
{
if(res3>=res1)
{
if(res2<=(res3+res5+res6-res1))
{
ans=res3+res5+res6;
}
else
{
ans=res1+res2;
}
}
else
{
ans=res3;
res1=res1-res3;
for(;res1>0;)
{
if(res5>0&&res1>=2)
{
res1=res1-2;
res5--;
ans=ans+2;
}
if(res1==1)
{
ans++;
res1=0;
}
if(res1==2&&res5==0)
{
ans=ans+2;
res1=0;
}
if(res6>0&&res1>=3)
{
res1=res1-3;
res6--;
ans=ans+3;
}
}
if(res2<=res6+res5)
{
ans=ans+res6+res5;
}
else
{
ans=ans+res2;
}
}
}
}
else
{
res1=0;
ans=max(res1,res3+res5+res6);
ans=max(ans,res2);
}
}
printf("%d\n",ans);
return 0;
}