Eugeny and Array（水题，注意题目描述即可）

最新推荐文章于 2025-04-07 21:51:44 发布

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最新推荐文章于 2025-04-07 21:51:44 发布

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原文链接：http://www.cnblogs.com/Staceyacm/p/10781943.html

本文介绍了一个算法问题，即如何响应针对由-1和1组成的数组的查询，判断是否可以通过重新排列数组元素使得指定区间内元素的和为0。文章提供了一段C++代码实现，通过计算数组中-1和1的数量来快速判断查询结果。

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

Query number i is given as a pair of integers li, ri (1 ≤ li ≤ ri ≤ n).
The response to the query will be integer 1, if the elements of array a can be rearranged so as the sum ali + ali + 1 + ... + ari = 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, ..., an (ai = -1, 1). Next m lines contain Eugene's queries. The i-th line contains integers li, ri (1 ≤ li ≤ ri ≤ n).

Output

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

Examples

Input

Output

0
1
0

Input

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

Output

代码

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>


using namespace std;

int main() {
	int n,m;
	int num[200005];
	cin>>n>>m;
	int sum1=0,sum2=0;
	for(int t=0; t<n; t++) {
		scanf("%d",&num[t]);
		if(num[t]==-1) {
			sum1++;
		} else {
			sum2++;
		}
	}
	int l,r;
	for(int t=0; t<m; t++) {
		scanf("%d%d",&l,&r);
		if((r-l)%2==0)
			printf("0\n");
		else if((r-l)%2!=0) {
			if((r-l+1)/2<=sum1&&(r-l+1)/2<=sum2)
				printf("1\n");
			else
				printf("0\n");
		}
	}



	return 0;
}

转载于:https://www.cnblogs.com/Staceyacm/p/10781943.html