Code Forces 313B Ilya and Queries

最新推荐文章于 2019-09-08 17:06:00 发布

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最新推荐文章于 2019-09-08 17:06:00 发布

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分类专栏：线性DP

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本文介绍了一个字符串匹配查询问题，给出了一段由字符“.”和“#”组成的字符串s及m个查询请求，每个查询请求包含两个整数li和ri，需要求出在范围(li, ri)内连续相同字符的数量。

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C - C

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 313B

Description

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 Input

Input

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

Output

Input

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

Output

DP。

dp[i]表示到这个i位置连续出现的字符数。

那么结果便是两头一减。

#include <stdio.h>
#include <string.h>
#define N 100005
char s[N];
int dp[N];
int main()
{
        int n,m;
        memset(dp,0,sizeof(dp));
        scanf("%s",s);
        n=strlen(s);
        scanf("%d",&m);
        for(int i=1;i<n;i++)
	{
		dp[i]+=dp[i-1];
		if(s[i]==s[i-1])
			dp[i]++;
	}
	int l,r;
        while(m--)
        {
              scanf("%d%d",&l,&r);
              printf("%d\n",dp[r-1]-dp[l-1]);
        }
        return 0;
}