Codeforces Round #186 (Div. 2) / 313B Ilya and Queries (字符串处理)

本文链接：https://blog.youkuaiyun.com/synapse7/article/details/10492401

本文介绍了一个字符串处理问题，通过一次遍历解决多个区间内的字符匹配计数问题，给出了一段高效简洁的 C++ 实现代码。

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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 "#" and m queries. Each query is described by a pair of integers l_i, r_i (1 ≤ l_i < r_i ≤ n). The answer to the query l_i, r_i is the number of such integers i (l_i ≤ i < r_i), that s_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 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 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 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

一遍扫描处理即可。

完整代码：

/*124ms,500KB*/

#include<cstdio>
#include<cstring>
const int maxn = 100010;

char str[maxn];
int count[maxn];

int main(void)
{
	gets(str);
	int len = strlen(str), m, l, r;
	for (int i = 1; i < len; ++i)
		count[i] = count[i - 1] + (str[i-1] == str[i]);
	scanf("%d", &m);
	while (m--)
	{
		scanf("%d%d", &l, &r);
		printf("%d\n", count[r - 1] - count[l - 1]);
	}
	return 0;
}