Code Forces 313B Ilya and Queries

C - C

Time Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

SubmitStatusPracticeCodeForces 313B

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

11220

---

DP。

dp[i]表示到这个i位置连续出现的字符数。

那么结果便是两头一减。

#include <stdio.h>#include <string.h>#define N 100005char s[N];int dp[N];int main(){        int n,m;        memset(dp,0,sizeof(dp));        scanf("%s",s);        n=strlen(s);        scanf("%d",&m);        for(int i=1;i<n;i++){dp[i]+=dp[i-1];if(s[i]==s[i-1])dp[i]++;}int l,r;        while(m--)        {              scanf("%d%d",&l,&r);              printf("%d\n",dp[r-1]-dp[l-1]);        }        return 0;}