关于openjudge 1.7节 34题回文字符串的解法

pineapple20 2025-02-04 13:20:34

'''

#include <cstdio>

#include <string>

#include <cctype>

#include <iostream>

using namespace std;

int main()

{

string s;

cin >> s;

int n = s.size(); // "China"

bool is_palindrome(string s);

for (int len= 2; len<=s.size(); len++)

for (int start = 0; start<=n - len; start++)

if ( is_palindrome(s.substr(start, len)) )

cout << s.substr(start, len) << endl;

return 0;

}

bool is_palindrome(string s)

{

int n = s.size(); // "abcba"

int i=0, j=n-1;

while (i<j)

if (s[i]==s[j])

{

i++;

j--;

}

else

return false;

return true;

}

'''

zhuyuhao

2025-02-04 13:23:32

//OpenJudge - 1.7 - 34 
#include <bits/stdc++.h>
using namespace std;
string oks[500];
bool cheak(string s){
	string hh = s;
	reverse(s.begin(),s.end());
	return hh == s;
}
bool cmp(string x,string y){
	return x.size() < y.size();
}
int main(){
	string s;
	cin>>s; 
	int n = s.size(),m = -1;
	if (n == 1){
		return 0;
	}
	for (int ok = 2;ok <= n;ok++){
		for (int i = 0;i <= n - ok;i++){
			string ook = s.substr(i,ok);
			if (cheak(ook)){
				cout<<ook<<endl;
			}
		}
	}
	return 0;
}

关于openjudge 1.7节 34题回文字符串的解法

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