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.
Examples
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
题目的意思是输入一行只有#和.的字符串，然后输入m对数（l，r）计算r-l中最大连续出现的字符有几个。
首先先计算字符串最大连续的字符有几个，记为数组dp，如果与前一个字符相同就加一，不同就等于上一次的dp。 #define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<string.h>
#define N 100005
char s[N];
int dp[N];
int main()
{
int l, r, i, m, n;
scanf("%s", s);
n = strlen(s);
memset(dp, 0, n);
scanf("%d", &m);
for (i = 1;i < n;i++)
{
dp[i] = dp[i - 1];
if (s[i] == s[i - 1])
dp[i]++;
}
for (i = 0;i < m;i++)
{
scanf("%d%d", &l, &r);
printf("%d
", dp[r - 1] - dp[l - 1]);
}
return 0;
}