标签：har cab possible hash else sof 组成 sample ==
题目描述
输入
输出
样例输入
Sample Input1:
7
ABXCABC
Sample Input2:
6
ABCDEF
Sample Input3:
9
ABABABABA
样例输出
Sample Output1:
ABC
Sample Output2:
NOT POSSIBLE
Sample Output3:
NOT UNIQUE
题解
字符串hash
首先偶数长度一定不存在，输出NOT POSSIBLE
奇数则先将每个前缀的hash值求出来，然后枚举多余字符位置，并根据位置与n/2和n/2+1的关系判断并求出前后两个字符串的hash值。
然后hash值相同时比较一下就好了。
#include <cstdio>
#define N 2000010
char str[N];
unsigned long long hash[N] , base[N] , x , y , ans;
int main()
{
int n , i , p , flag = 0 , sum = 0;
scanf("%d%s" , &n , str + 1);
if(n % 2 == 0)
{
printf("NOT POSSIBLE\n");
return 0;
}
base[0] = 1;
for(i = 1 ; i <= n ; i ++ ) hash[i] = hash[i - 1] * 131 + str[i] , base[i] = base[i - 1] * 131;
for(i = 1 ; i <= n ; i ++ )
{
if(i <= n / 2) x = hash[n / 2 + 1] - hash[i] * base[n / 2 - i + 1] + hash[i - 1] * base[n / 2 - i + 1];
else x = hash[n / 2];
if(i <= n / 2 + 1) y = hash[n] - hash[n - n / 2] * base[n / 2];
else y = (hash[i - 1] - hash[n / 2] * base[i - n / 2 - 1]) * base[n - i] + hash[n] - hash[i] * base[n - i];
if(x == y)
{
if(flag && ans != x)
{
printf("NOT UNIQUE\n");
return 0;
}
flag = 1 , ans = x , p = i;
}
}
if(!flag)
{
printf("NOT POSSIBLE\n");
return 0;
}
for(i = 1 ; sum < n / 2 ; i ++ )
if(i != p)
printf("%c" , str[i]) , sum ++ ;
printf("\n");
return 0;
}
【bzoj3916】[Baltic2014]friends 字符串hash
标签：har cab possible hash else sof 组成 sample ==
原文地址：http://www.cnblogs.com/GXZlegend/p/6871534.html
