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CodeForces 688B - Lovely Palindromes(思路)

题意:输出第n(1 <= n <= 10^100000)大的偶数长度的回文数。(最小的为11)

因为长度是偶数,所以前后两半之间是相互对称的,又因为一个数字的大小主要取决于较高位数的大小,所以数字的前一半决定数的大小,从1开始,1,2,3……对称即可得11,22,33……

所以将数正着输出后再倒着输出一遍即可(数非常大,需用字符串存)

 

#include<cstdio>  
#include<cstring>  
#include<cctype>  
#include<cstdlib>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<deque>  
#include<queue>  
#include<stack>  
#include<list>  
typedef long long ll;  
typedef unsigned long long llu;  
const int MAXN = 100 + 10;  
const int MAXT = 1000000 + 10;  
const int INF = 0x7f7f7f7f;  
const double pi = acos(-1.0);  
const double EPS = 1e-6;  
using namespace std;  
  
string s;  
  
int main(){  
    cin >> s;  
    cout << s;  
    reverse(s.begin(), s.end());  
    cout << s << endl;  
    return 0;  
}  

 

CodeForces 688B - Lovely Palindromes(思路)