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

......
4
3 4
2 3
1 6
2 6

output

1
1
5
4

input

#..###
5
1 3
5 6
1 5
3 6
3 4

output

1
1
2
2
0

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;
}