Code Forces 313B Ilya and Queries

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

DP.
dp[i]는 이 i 위치에서 연속적으로 나타나는 문자 수를 나타낸다.
그럼 결과는 양쪽에서 하나씩 빼는 거야.

#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
",dp[r-1]-dp[l-1]);
        }
        return 0;
}