题目位置

最大回文数(number)

解释代码

一句话,想说的全在代码中...

代码

#include <bits/stdc++.h>
using namespace std;

//这道题我开始用的long long可只得了60分,于是我另辟蹊径运用了string最终成功AC

bool dy (string a, string b) {
	if(a.length() > b.length()) {
		return true;
	}
	if(a.length() < b.length()) {
		return false;
	}
	return a > b;
}

int main() {
	freopen("number.in", "r", stdin);
	freopen("number.out", "w", stdout);//这是很重要的
	long long n;
	string a, b, c = "";
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a;
		b = a;
		reverse(b.begin(), b.end());//reverse将字符串倒转
		if (a == b) { //判断是否一样
			if(dy(a, c)) {
				c = a;
			}
		}
		b = ""; 
	}
	cout << c << endl;
	/*
		fclose(stdin);
		fclose(stdout);
	*/
	//以上两条可写可不写!!
	return 0;
}