cf 186 div2 B

本博客探讨了如何解决字符串中特定字符在不同区间内的出现次数问题,包括输入字符串和一系列查询,通过预处理优化查询效率。

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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
代码:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>

using namespace std;

const int maxn=100005;
char str[maxn];
int ans[maxn];
int m;
int l,r;

int main(){
    int n,i,j;
    cin>>str;
    n=strlen(str);
    ans[0]=0;
    for(int i=1;i<n;i++){
        if(str[i]==str[i-1])
           ans[i]=ans[i-1]+1;
        else
           ans[i]=ans[i-1];
    }
    cin>>m;
    while(m--){
         cin>>l>>r;
         printf("%d\n",ans[r-1]-ans[l-1]);;
    }
    return 0;
}


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