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  • Eugeny and Array(水题,注意题目描述即可)

    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 liri (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

    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

    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
    ");
    		else if((r-l)%2!=0) {
    			if((r-l+1)/2<=sum1&&(r-l+1)/2<=sum2)
    				printf("1
    ");
    			else
    				printf("0
    ");
    		}
    	}
    
    
    
    	return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10781943.html
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