a1Ex999 @ 2024-10-25 21:50:59
#include <bits/stdc++.h>
using namespace std;
int some=0;
bool ask=0;
string s;
int main(){
cin>>s;
for(int i=0;i<s.size();i++){
ask=0;
if(s[i]==','||i==s.size()-1){
if(i-some>12||i-some<6){
some=i;
}else{
for(int j=some;j<=i;j++){
if(s[j]>='a'&&s[j]<='z') continue;
if(s[j]>='A'&&s[j]<='Z') continue;
if(s[j]>='0'&&s[j]<='9') continue;
if(s[j]=='!') continue;
if(s[j]=='@') continue;
if(s[j]=='#') continue;
if(s[j]=='$') continue;
if(s[j]==',') continue;
ask=1;
}
if(ask==0){
for(int j=some;j<=i;j++){
if(s[j]!=',') cout<<s[j];
}
cout<<endl;
some=i;
}
}
}
}
return 0;
}by wandereman @ 2024-10-25 22:22:29
你要把每种数量都计算出来，否则可能出现在一个密码中只出现了一种或两种，但你的程序不会把它排除掉，样例能过是因为后面两个的长度都大于12.
by wandereman @ 2024-10-25 22:24:44
而你的程序只能保证都是合法的（没有像&*…%这种）
by wandereman @ 2024-10-25 22:29:14
看一下吧```cpp
using namespace std; string s; int main() { cin>>s; string t=""; for(int i=0;i<s.length();i++) { if (s[i]!=',') { t += s[i]; } else { bool pd=1; int xx=0,dx=0,sz=0,tsfh=0; for(int j=0;j<t.length();j++) { if(t[j]>='a'&&t[j]<='z') { xx=1; } else if(t[j]>='A'&&t[j]<='Z') { dx=1; } else if(t[j]>='0'&&t[j]<='9') { sz=1; } else if (t[j]=='!'||t[j]=='@'||t[j]=='#'||t[j]=='$') { tsfh=1; } else { pd=0; break; } } if(pd&&xx+dx+sz>=2&&tsfh>=1&&tsfh+xx+dx+sz>=6&&tsfh+xx+dx+sz<=12) { cout<<t<<endl; } t=""; } } return 0; }
by wandereman @ 2024-10-25 22:29:36
求关
by wandereman @ 2024-10-25 22:31:34
看一下吧ACcode
by a1Ex999 @ 2024-11-24 15:19:20
@wandereman