題干：

Mike has a string?s?consisting of only lowercase English letters. He wants to?change exactly one?character from the string so that the resulting one is a palindrome.

A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.

Input

The first and single line contains string?s?(1?≤?|s|?≤?15).

Output

Print "YES" (without quotes) if Mike can change?exactly?one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.

Examples

Input

abccaa

Output

YES

Input

abbcca

Output

NO

Input

abcda

Output

YES

題目大意：

麥克有一個只包含小寫字母的字符串?s?。他想知道他能否恰好改變這個字符串的一個字母使得這個字符串成為回文串。

回文串是從前向后讀與從后向前讀相同的串，舉個例子， "z", "aaa", "aba", "abccba" 都是回文串，但字符串 "codeforces", "reality", "ab" 就不是。

解題報告：

? 分奇偶數(shù)討論就行了，，注意是恰好改變一個字符，，也就是說不能不改。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; char s[MAX]; int main() {cin>>(s+1);int len = strlen(s+1);int cnt = 0;if(len&1) {for(int i = 1; i<=(len>>1); i++) {if(s[i] != s[len-i+1]) cnt++;}if(cnt <= 1) puts("YES");else puts("NO");}else {for(int i = 1; i<=(len>>1); i++) {if(s[i] != s[len-i+1]) cnt++;}if(cnt == 1) puts("YES");else puts("NO");}return 0 ;}

很早之前寫的垃圾代碼：（我當(dāng)時也太水了吧）

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; char a[20]; char b[20]; int main() {cin>>a;int len=strlen(a);int flag=0; strcpy(b,a);reverse(b,b+len); // printf("%d\n",len);for(int i = 0; i<len; i++) {if(a[i]!=b[i]) {b[i]=a[i];a[len-1-i]=b[i];flag=1;break;}} // printf("%s\n%s",a,b);if(len&1) {for(int i = 0; i<len; i++) {if(a[i]!=b[i]) {printf("NO");return 0 ;}}printf("YES");return 0 ; }else {if(flag==0) {printf("NO");return 0 ;}for(int i = 0; i<len; i++) {if(a[i]!=b[i]) {printf("NO");return 0 ;}}printf("YES");return 0 ; } }

?

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