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

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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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

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;
}


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