Subpalindromes

Subpalindromes

Time Limit: 500ms

Memory Limit: 65536KB

This problem will be judged on Ural. Original ID:  1989
64-bit integer IO format:  %lld      Java class name:  (Any)

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You have a string and queries of two types:

1. replace i’th character of the string by character a;
2. check if substring s_j...s_k is a palindrome.

Input

The first line contains a string consisting of  n small English letters. The second line contains an integer  m that is the number of queries (5 ≤  n,  m ≤ 10 ⁵). The next m lines contain the queries. 
Each query has either form “change i a”, or “palindrome? j k”, where  i,  j,  k are integers (1 ≤  i ≤  n; 1 ≤  j ≤  k ≤  n), and character  a is a small English letter.

Output

To all second type queries, you should output “Yes” on a single line if substring  s _j... s _k is a palindrome and “No” otherwise.

Sample Input

input	output
abcda 5 palindrome? 1 5 palindrome? 1 1 change 4 b palindrome? 1 5	No Yes Yes Yes

题意：给一个字符串，进行两个操作：

            1.change i c ，把第i个换成c；

            2.palindrome？ i j ，询问第i个字符到第j个字符是不是回文串？

代码如下：

#include<stdio.h>
#include<string.h>
#define maxn 1100100

char str[maxn];
long long kk[maxn];

struct Node
{
	int l,r;
	long long s,f; 
}node[maxn*4];

void hashinist()
{
	int i;
	kk[0]=1;
	for(i=1;i<maxn;i++) kk[i]=kk[i-1]*27;    //一开始乘了26，wa了......
}

Node cnt(Node a,Node b)
{
	Node ans;
	ans.l=a.l;
	ans.r=b.r;
	ans.s=a.s*kk[b.r-b.l+1]+b.s;      //从l到r的字符串hash一遍
	ans.f=b.f*kk[a.r-a.l+1]+a.f;      //从r到l的字符串hash一遍
	return ans;
}

void build(int i,int l,int r)
{
	node[i].l=l;
	node[i].r=r;
	if(l==r)
	{
		node[i].s=node[i].f=str[l-1]-'a';
		return ;
	}
	int mid=(l+r)/2;
	build(i*2,l,mid);
	build(i*2+1,mid+1,r);
	node[i]=cnt(node[i*2],node[i*2+1]);
}

void update(int i,int j,int w)
{
	if(node[i].l==node[i].r&&node[i].l==j)
	{
		node[i].s=node[i].f=w;
		return ;
	}
	int mid=(node[i].l+node[i].r)/2;
	if(j<=mid) update(i*2,j,w);
	else update(i*2+1,j,w);
	node[i]=cnt(node[i*2],node[i*2+1]);
}

Node query(int i,int l,int r)
{
	if(node[i].l==l&&node[i].r==r) return node[i];
	int mid=(node[i].l+node[i].r)/2;
	if(mid>=r) return query(i*2,l,r);
	else if(mid<l) return query(i*2+1,l,r);
	else
	{
		Node ans;
		ans=cnt(query(i*2,l,mid),query(i*2+1,mid+1,r));
		return ans;
	}
}

int main()
{
	while(scanf("%s",str)!=EOF)
	{
		int n;
		hashinist();
		int l=strlen(str);
		build(1,1,l);
		scanf("%d",&n);
		while(n--)
		{
			char z[50];
			scanf("%s",z);
			if(z[0]=='c')
			{
				int w;
				char c;
				scanf("%d %c",&w,&c);
				update(1,w,c-'a');
			}
			else
			{
				int u,v;
				scanf("%d %d",&u,&v);
				Node ans;
				ans=query(1,u,v);
				if(ans.f==ans.s) puts("Yes");
				else puts("No");
			}
		}
	}
	return 0;
}