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
題目的意思是輸入一行隻有#和.的字元串，然後輸入m對數（l，r）計算r-l中最大連續出現的字元有幾個。
首先先計算字元串最大連續的字元有幾個，記為數組dp，如果與前一個字元相同就加一，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;
}