AC dreamoj 1011 树状数组+hash维护字符串的前缀和

http://acdream.info/problem?pid=1019

Problem Description

Now we have a long long string, and we will have two kinds of operation on it.

C i y : change the ith letter to y.

Q i j : check whether the substring from ith letter to jth letter is a palindrome.

Input

There are multiple test cases.

The first line contains a string whose length is not large than 1,000,000.

The next line contains a integer N indicating the number of operations.

The next N lines each lines contains a operation.

All letters in the input are lower-case.

Output

For each query operation, output "yes" if the corresponding substring is a palindrome, otherwise output "no".

Sample Input

aaaaa
4
Q 1 5
C 2 b
Q 1 5
Q 1 3

Sample Output

yes
no
yes

/**
ACdreamoj 1019  树状数组+hash维护字符串的前缀和
题目大意：
         给定一个字符串，单点更新，区间查询该区间是不是回文串。
解题思路：
         hash是x1 * p^1+ x2*p^2 +x3*p^3...可以用树状数组维护前缀和，
维护两个串，一个是正串，另一个是反串用于比较。
字符串区间s[l~r]的哈希值为sum(s[r]-s[l-1])/Hash[l-1];
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef unsigned long long LL;
const int maxn=1000010;
const int seed=13;

LL Hash[maxn],C[2][maxn];
char s[maxn];
int len;

void init()
{
    Hash[0]=1;
    for(int i=1;i<maxn;i++)
         Hash[i]=Hash[i-1]*seed;
}

int lowbit(int x)
{
    return x&(-x);
}

void add(int i,int x,LL pos)
{
    while(x<=len)
    {
        C[i][x]+=pos;
        x+=lowbit(x);
    }
}

LL sum(int i,int x)
{
    LL ans=0;
    while(x)
    {
        ans+=C[i][x];
        x-=lowbit(x);
    }
    return ans;
}
LL gethash(int i,int l,int r)
{
    return sum(i,r)-sum(i,l-1);
}
int main()
{
     init();
     while(~scanf("%s",s+1))
     {
         memset(C,0,sizeof(C));
         len=strlen(s+1);
         for(int i=1;i<=len;i++)
         {
             add(0,i,(s[i]-'a')*Hash[i]);
             add(1,len+1-i,(s[i]-'a')*Hash[len+1-i]);
         }
         int T;
         cin >> T;
         while(T--)
         {
             char c[5];
             scanf("%s",c);
             if(c[0]=='C')
             {
                 char b[5];
                 int a;
                 scanf("%d%s",&a,b);
                 add(0,a,(b[0]-s[a])*Hash[a]);
                 add(1,len+1-a,(b[0]-s[a])*Hash[len+1-a]);
                 s[a]=b[0];
             }
             else
             {
                 int l,r;
                 scanf("%d%d",&l,&r);
                 if(gethash(0,l,r)*Hash[len-r]==gethash(1,len+1-r,len+1-l)*Hash[l-1])//采用交叉相乘把除换成了乘
                      puts("yes");
                 else
                      puts("no");
             }
         }
     }
     return 0;
}