CF:Ilya and Queries

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), that si = 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

解题报告：大概意思是先给出一个串s，然后给出m行，每行给出l,r两个数求满足s[i+1]==s[i];的个数.l<=i<r;

         可以定义一个整形数组A【】;然后求出每一位之前符合要求的个数，当输入了l,r；时，A[r]-A[l]即为【l,r】之间符合的个数；

参考代码：

#include <stdio.h>#include<string.h>char str[100010];int  A[100010];int main(){//freopen("in.txt","r",stdin);int i,n,m,j,k,t,a,b,s;scanf("%s",str);memset(A,0,sizeof(A));scanf("%d ",&m);n=strlen(str);for(i=1;i<n;i++){A[i]=A[i-1]+(str[i]==str[i-1]);       }while(m--){ s=0;  scanf("%d %d ",&a,&b);     printf("%d\n",A[b-1]-A[a-1]);   }return 0;}