【Gym 100971K】Palindromization
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.
The input contains a string s of length (2 ≤ |s| ≤ 200000), consisting of lowercase Latin letters.
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).
evertree
YES
2
emerald
NO
aa
YES
2
从左右两边往中间移动,如果相同,l++,r--,如果不同,第一次循环i==0时,则让l++,第二次循环则让r--,最后判断不同的出现了几次。如果dif是0,l>=r代表本身就是回文,则去掉l位置的字符。
#include <cstring> #include <cstdio> #define N 100005 char s[N<<1]; int len,l,r,dif,at; int main() { scanf("%s",s); len=strlen(s); for(int i=0;i<2;i++){ dif=0; l=0,r=len-1; while(l<r){ if(s[l]!=s[r]){ dif++; if(i){ at=r; r--; }else { at=l; l++; } } if(s[l]==s[r]){ l++; r--; } } if(l>=r&&dif==0)at=l; if(dif<2){ printf("YES\n%d\n",at+1); break; } } if(dif>1)puts("NO"); }
┆凉┆暖┆降┆等┆幸┆我┆我┆里┆将┆ ┆可┆有┆谦┆戮┆那┆ ┆大┆始┆ ┆然┆
┆薄┆一┆临┆你┆的┆还┆没┆ ┆来┆ ┆是┆来┆逊┆没┆些┆ ┆雁┆终┆ ┆而┆
┆ ┆暖┆ ┆如┆地┆站┆有┆ ┆也┆ ┆我┆ ┆的┆有┆精┆ ┆也┆没┆ ┆你┆
┆ ┆这┆ ┆试┆方┆在┆逃┆ ┆会┆ ┆在┆ ┆清┆来┆准┆ ┆没┆有┆ ┆没┆
┆ ┆生┆ ┆探┆ ┆最┆避┆ ┆在┆ ┆这┆ ┆晨┆ ┆的┆ ┆有┆来┆ ┆有┆
┆ ┆之┆ ┆般┆ ┆不┆ ┆ ┆这┆ ┆里┆ ┆没┆ ┆杀┆ ┆来┆ ┆ ┆来┆
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