[Codeforces Round #186 (Div. 2)] B. Ilya and Queries

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), 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

题解：一个字符串只有.或者#字符，给你一段区间[x,y]，y>x，在这个区间里面统计s[i]=s[i+1]最多个数。
线性动态规划：转移方程

　    　  if(a[i]==a[i-1])
　　　　f[i]=f[i-1]+1;
　　　　else f[i]=f[i-1];

 题目地址：http://codeforces.com/contest/313/problem/B

代码：

 

#include<stdio.h>
#include<string.h>

int i,j,n,x,y,m,
    f[110000];
char a[110000];


int
main()
{
    gets(a);
    m=strlen(a);
    f[0]=0;
    for(i=1;i<m;i++)
    {
        if(a[i]==a[i-1])
        f[i]=f[i-1]+1; 
        else f[i]=f[i-1];
    }
   
   scanf("%d",&n);
   while(n--)
    {
        scanf("%d%d",&x,&y);
        printf("%d\n",f[y-1]-f[x-1]);
    } 
    
    return 0;
}