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

aaaaa4Q 1 5C 2 bQ 1 5Q 1 3

Sample Output

yesnoyes

/**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;}