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

......43 42 31 62 6

output

1154

input

#..###51 35 61 53 63 4

output

11220

解題說明：此題是求一個字串中一段指定位置內連續出現的相同的字元有多少個，最簡單的的想法就是針對每一次尋找遍曆相應位置輸出結果，但是在字串較長時肯定會逾時。為此，我們應該在輸入字串後就進行處理，用一個和字串長度相同的int數組儲存到字串這個位置連續出現的字元數有多少，這一點和動態規劃的思想類似。最後查詢時只需要用末尾的int值減去開頭的int值即可，這個增加的值就是在這段區間內連續出現的字元數目。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <algorithm>using namespace std;int main(){char s[100009];int a[100009]={0},i,j,n,x,y,l;scanf("%s",s);l=strlen(s);for(i=1;i<l;i++){a[i]+=a[i-1];if(s[i]==s[i-1]){a[i]++;}}scanf("%d",&n);while(n--){scanf("%d%d",&x,&y);printf("%d\n",a[y-1]-a[x-1]);}return 0;}