ACM学习感悟——weekly training

Description

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...s_n (n is the length of the string), consisting only of characters "." and "#" andm queries. Each query is described by a pair of integersl_i, r_i(1 ≤ l_i < r_i ≤ n). The answer to the query l_i, r_i is the number of such integersi(l_i ≤ i < r_i), thats_i = s_i + 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 lengthn(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 nextm lines contains the description of the corresponding query. Thei-th line contains integers l_i, r_i(1 ≤ l_i < r_i ≤ n).

Output

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

Sample Input

Input

......43 42 31 62 6

Output

1154

Input

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

Output

1

1

2

2

0

这是动态规划的题目，难度一般。状态转移方程：d[i]=s[i]==s[i-1]?d[i-1]+1:d[i]。

CODE：

///////////////////////////////////////////////////////// //                           //                        ////  Created by 吴尔立                        ////  Copyright (c) 2015年 吴尔立. All rights reserved.  ///////////////////////////////////////////////////////////#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>           #include <algorithm>#include <cctype>#include <stack>#include <queue>#include <map>#include <string>#include <set>#include <vector>#define ll long long;#define INF 1<<31#define cir(i,a,b)  for (int i=a;i<=b;i++)#define CIR(j,a,b)  for (int j=a;j>=b;j--)#define CLR(x) memset(x,0,sizeof(x))using namespace std;#define maxn 100005long long n,m;string s;long long l[maxn],r[maxn];long long d[maxn];int main(){getline(cin,s);n=s.length();scanf("%ld",&m);for (long long i=1;i<=m;i++){scanf("%ld%ld",&l[i],&r[i]);}//for (long long i=0;i<n;i++) cout << s[i] << endl;//cout << m << endl; CLR(d);for (long long i=1;i<n;i++){if (s[i]==s[i-1]) d[i]=d[i-1]+1;else d[i]=d[i-1];}for (long long i=1;i<=m;i++)cout << d[r[i]-1]-d[l[i]-1] << endl;return 0;}