Codeforces Round #182 (Div. 2) / 302A Eugeny and Array(模拟)_c语言 acm there is an array with n elements a1 a2 ..-优快云博客

本文介绍 CodeForces 平台上题目 302A 的解题思路与实现方法，该题涉及数组操作及区间查询，通过判断是否能重新排列数组中的元素使指定区间的和为零。

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Eugeny has array a = a₁, a₂, ..., a_n, consisting of n integers. Each integer a_i equals to -1, or to 1. Also, he has m queries:

Query number i is given as a pair of integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n).
The response to the query will be integer 1, if the elements of array a can be rearranged so as the sum a_{l_i} + a_{l_i + 1} + ... + a_{r_i} = 0, otherwise the response to the query will be integer 0.

Help Eugeny, answer all his queries.

Input

The first line contains integers n and m (1 ≤ n, m ≤ 2·10⁵). The second line contains n integers a₁, a₂, ..., a_n (a_i = -1, 1). Next mlines contain Eugene's queries. The i-th line contains integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n).

Output

Print m integers — the responses to Eugene's queries in the order they occur in the input.

   Sample test(s)
 

     input
   
2 3
1 -1
1 1
1 2
2 2

     output
   
0
1
0

     input
   
5 5
-1 1 1 1 -1
1 1
2 3
3 5
2 5
1 5

     output
   
0
1
0
1
0

写第一遍没看清题意，样例2看不懂，后来发现我确实要补补英语了。。（不过这句话确实很容易漏掉）

原文有这么一句话：

" if the elements of array a can be rearranged ..."（如果序列a中元素经过重新排列后…）

完整代码：

/*171ms,0KB*/

#include<cstdio>

int main()
{
	int n, m, a;
	int c1 = 0, c2 = 0;
	int l, r;
	scanf("%d%d", &n, &m);
	while (n--)
	{
		scanf("%d", &a);
		if (a > 0)
			++c1;
		else
			++c2;
	}
	while (m--)
	{
		scanf("%d%d", &l, &r);
		a = r - l + 1;///第一次交的时候这里写反了，WA4，罪过，罪过。。。
		if (a & 1)
			puts("0");
		else
		{
			a >>= 1;
			puts(c1 < a || c2 < a ? "0" : "1");
		}
	}
	return 0;
}