CF_313B_IlyaAndQueries

B. Ilya and Queries

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Ilya the Lion wants to help all his friends with passing exams. They need to solve the following problem to pass the IT exam.

You've got string s = s1s2... sn (n is the length of the string), consisting only of characters "." and "#" and m queries. Each query is described by a pair of integers li, ri (1 ≤ li < ri ≤ n). The answer to the query li, ri is the number of such integers i (li ≤ i < ri), thatsi = si + 1.

Ilya the Lion wants to help his friends but is there anyone to help him? Help Ilya, solve the problem.

Input

The first line contains string s of length n (2 ≤ n ≤ 105). It is guaranteed that the given string only consists of characters "." and "#".

The next line contains integer m (1 ≤ m ≤ 105) — the number of queries. Each of the next m lines contains the description of the corresponding query. The i-th line contains integers li, ri (1 ≤ li < ri ≤ n).

Output

Print m integers — the answers to the queries in the order in which they are given in the input.

Sample test(s)

input

......43 42 31 62 6

output

1154

input

#..###51 35 61 53 63 4

output

11220

最开始一看题目，区间和，马上想到树状数组

但是树状数组也打完了，反应过来，这根本没有数据修改的

根本用不到树状数组

直接扫一遍求和就可以了

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int M=1e5+5;char s[M];int a[M];int main(){    int nq;    scanf("%s",s);    scanf("%d",&nq);    int len=strlen(s);    for(int i=0;i<len;i++)    {        if(s[i]==s[i+1])            a[i+1]+=a[i]+1;        else            a[i+1]=a[i];    }    int aa,b;    for(int i=1;i<=nq;i++)    {        scanf("%d%d",&aa,&b);        printf("%d\n",a[b-1]-a[aa-1]);    }    return 0;}