您的位置:首页 > 其它

codeforces gym 100971 K Palindromization 思路

2017-08-27 21:59 531 查看
题目链接:http://codeforces.com/gym/100971/problem/K

K. Palindromization

time limit per test
2.0 s

memory limit per test
256 MB

input
standard input

output
standard output

Mihahim has a string s. He wants to delete exactly one character from it so that the resulting string would be a palindrome. Determine if he can do it, and if he can, what character should be deleted.

Input
The input contains a string s of length (2 ≤ |s| ≤ 200000), consisting of lowercase Latin letters.

Output
If the solution exists, output «YES» (without quotes) in the first line. Then in the second line output a single integer x — the number of the character that should be removed from s so that the resulting string would be a palindrome. The characters in the string are numbered from 1. If there are several possible solutions, output any of them.

If the solution doesn't exist, output «NO» (without quotes).

Examples

input
evertree


output
YES
2


input
emerald


output
NO


input
aa


output
YES
2

题意:给你一个字符串,删除一个字符,是否形成回文,如果有,输出删除的位置;

思路:对于删除一个字符以后,这个字符串的对称轴的位置只有两个,所有枚举对称轴,复杂度O(n);

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
#define LL long long
const int N=2e5+100,M=1e6+10,inf=1e9+10;
const LL INF=1e18+10,mod=19260817;

char a
;
int checkl(int l,int r,int n,int L)
{
while(r<=n)
{
while(a[l]!=a[r])
{
if(L!=1)return 0;
L=l;
l--;
}
l--;r++;
}
return L;
}
int checkr(int l,int r,int n,int R)
{
while(l>=1)
{
while(a[l]!=a[r])
{
if(R!=n)return 0;
else R=r;
r++;
}
l--;r++;
}
return R;
}
int main()
{
scanf("%s",a+1);
int n=strlen(a+1);
if(n%2==0)
{
int ans=checkl(n/2,n/2+2,n,1);
if(ans)return 0*printf("YES\n%d\n",ans);
ans=checkr(n/2-1,n/2+1,n,n);
if(ans)return 0*printf("YES\n%d\n",ans);
puts("NO");
}
else
{
int ans=checkl(n/2+1,n/2+2,n,1);
if(ans)return 0*printf("YES\n%d\n",ans);
ans=checkr(n/2,n/2+1,n,n);
if(ans)return 0*printf("YES\n%d\n",ans);
puts("NO\n");
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: