UVA - 10324 Zeros and Ones

Description

Given a string of 0's and
1's up to
1000000 characters long and indices
i and
j, you are to answer a question whether all characters between position
min(i,j) and position
max(i,j) (inclusive) are the same.

Input

There are multiple cases on input. The first line of each case gives a string of 0's and 1's. The next line contains a positive integer n giving the number of queries for this case. The next n lines
contain queries, one per line. Each query is given by two non-negative integers, i and j. For each query, you are to print Yes if all characters in the string between position min(i,j) and
position max(i,j) are the same, and No otherwise.

Output

Each case on output should start with a heading as in the sample below. The input ends with an empty string that is a line containing only the new line character, this string should not be processed. The input
may also with end of file. So keep check for both.

Sample Input

0000011111
3
0 5
4 2
5 9
01010101010101010101010101111111111111111111111111111111111110000000000000000
5
4 4
25 60
1 3
62 76
24 62
1
1
0 0

Sample Output

Case 1:
No
Yes
Yes
Case 2:
Yes
Yes
No
Yes
No
Case 3:
Yes

题意：给你一个01串，推断连续的字串[l, r]是否仅仅含同一个种字符

思路：用dp[i]记录到当前位置有多少个连续的同样字符，假设询问的区间的大小<=dp[r],那么代表就是能够的

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 1000005;

int dp[maxn];
int n, q;
char str[maxn];

int main() {
	int cas = 1;
	while (scanf("%s", str) != EOF) {
		int len = strlen(str);
		dp[0] = 1;
		for (int i = 1; i < len; i++)
			if (str[i] == str[i-1])
				dp[i] = dp[i-1] + 1;
			else dp[i] = 1;
		printf("Case %d:\n", cas++);
		int l, r;
		scanf("%d", &q);
		while (q--) {
			scanf("%d%d", &l, &r);
			if (l > r)
				swap(l, r);
			if (r - l + 1 <= dp[r])
				printf("Yes\n");
			else printf("No\n");
		}
	}
	return 0;
}