技術標籤：題解hash

思路：

首先我們把字串存到hash陣列當中，然後思考怎樣判斷
我們可以列舉中心位置，然後二分長度，用字串hash判斷是否對稱就行
注意要判斷長度奇偶

c o d e code code

#include<iostream>
#include<cstdio>
#include<cstring> 
using namespace std;
char s[1000100];
unsigned long long up[1000100], down[1000100], f[1000100];
int main()
{
	scanf("%s", s+
 
1);
	int n=strlen(s+1);
	int m=0;
	while(s[1]!='E'||s[2]!='N'||s[3]!='D')
	{
		f[0]=1;
		for(int i=1; i<=n; i++)
		{
			f[i]=f[i-1]*131ull;
			up[i]=up[i-1]*131ull+s[i]-'a';
		}
		down[n+1]=0;
		for(int i=n; i>=1; i--)
			down[i]=down[i+1]*131ull+s[i]-'a';
		int ans=0;
		for(int i=1; i<=n; i++)
 

		{
			int l=0, r=n;
			while(l<=r)
			{
				int mid=l+r>>1;
				if(i-mid<1||i+mid>n)
					r=mid-1;
				else
				if(up[i]-up[i-mid-1]*f[mid+1]==down[i]-down[i+mid+1]*f[mid+1])
					ans=max(ans, mid*2+1), l=mid+1;
				else
					r=mid-1;
			}
			l=0, r=n;
			while(l<=r)
			{
				int mid=
 
l+r>>1;
				if(i-mid<0||i+mid>n)
					r=mid-1;
				else
				if(up[i]-up[i-mid]*f[mid]==down[i+1]-down[i+mid+1]*f[mid])
					ans=max(ans, mid*2), l=mid+1;
				else
					r=mid-1;
			}
		}
		m++;
		printf("Case %d: %d\n", m, ans);
		memset(down, 0, sizeof(down));
		scanf("%s", s+1);
		n=strlen(s+1);
	}
	return 0;
}
/*abcbabcbabcba*/