A. Eugeny and Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)input

2 31 -11 11 22 2

output

010

input

5 5-1 1 1 1 -11 12 33 52 51 5

output

01010

解題說明：題目中給定一個只由-1,1組成的數列，在給定起始位置和結束位置後，問在可以對數列中數字進行任意排列的情況下能否讓從給定的起始位置到結束位置之間的數字之和為0. 做法是先統計出數列中1和-1的個數，然後判斷起始位置和結束位置中間包含多少個數字，如果為奇數那和肯定不為0，如果為偶數就判斷數列中1和-1的個數能否達到該數的一半，只有一半1和一半-1時才能確保加起來為0.

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <algorithm>using namespace std;int main(){int i,n,m;int l,r;int a,pos,neg;pos=neg=0;scanf("%d %d",&n,&m);for(i=0;i<n;i++){scanf("%d",&a);if(a>0){pos++;}else{neg++;} } if(pos>neg) { pos=neg; } for(i=0;i<m;i++) {scanf("%d %d",&l,&r);l=r-l+1; if (l%2==1){printf("0\n");}else if (pos>=l/2){printf("1\n");}else{printf("0\n");} } return 0;}