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Time Limit: 20 Sec
Memory Limit: 256 MB
http://codeforces.com/gym/100187/problem/K
Andrey works in the Ministry of Truth. His work is changing articles in newspapers and magazines so that they praise the Party and Big Brother.
Recently Big Brother decided that it would be fine if all words in all articles are read equally both from right to left and from left to right. In his opinion it will greatly simplify the reading of articles — even if the word is read in the reverse order, its sense won‘t change.
Andrey spends one second to erase one letter from the word and write another one to its place. Only one word left to be changed, after that he will complete the plan and be able to go home. Of course, he wants to do it as soon as possible. But the truth is that he does not clearly understand right now which word he should get after the replacement. Help him to find it out.
Input
Input contains a single line with the word that Andrey has to change. The word consists of lowercase Latin letters and its length is from 1 to 200000, inclusively.
Output
Output the word which should be a result of Andrey‘s work. If there are several possible such words, output any of them.
Sample Input
abccabd
Sample Output
abacaba
题意
花费最少，构造回文
题解：
水题系列……
代码
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 2000001 #define mod 10007 #define eps 1e-9 const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } //************************************************************************************** int main() { string s; cin>>s; for(int i=0;i<s.size();i++) if(s[i]!=s[s.size()-1-i]) s[s.size()-1-i]=s[i]; cout<<s<<endl; }
codeforces Gym 100187L L. Ministry of Truth 水题
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原文地址：http://www.cnblogs.com/qscqesze/p/4657736.html
