A.Eugeny and Array

              A. Eugeny and Array

Time Limit:2000ms
Case Time Limit:2000ms
Memory Limit:262144KB

64-bit integer IO format:%I64d      Java class name:(Any)
SubmitStatusPID:
29452
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Eugeny has arraya = a1, a2, ..., an,
consisting ofn integers. Each integer
ai equals to -1, or to 1. Also, he hasm queries:
Query number i is given as a pair of integersli,ri(1 ≤ li ≤ ri ≤ n).
The response to the query will be integer
1, if the elements of arraya can berearranged(重新排列) so as the sumali + 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 containsn integersa1, a2, ..., an(ai = -1, 1).
Nextm lines contain Eugene's queries. Thei-th line contains integersli, ri(1 ≤ li ≤ ri ≤ n).

Output

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

Sample Input

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
01
0

解题思路：本题为简单题，只要搞懂题目，就能做出来。

给你一个数列，数组元素为1或-1，要你对数列进行重排，看能否满足给定的标号l和r使得数列中l~r之间的数的和为0，若可以，输出1，否则输出0。

只要记录数列中1，-1的个数然后根据给定的l和r判断即可。若l~r区间中的数正好有偶数个数，而给定的数列中1，和-1的个数都不小于l~r区间中的数的个数的1/2，那么，我们就可以重排序列使其满足要求。如若这样，输出1，否则，输出0。

#include<stdio.h>
int main()
{
int n,m;
int x,y,a,b,i;
while(scanf("%d%d",&n,&m)!=EOF)
{
a=0;
b=0;
for(i=0; i<n; i++)
{
scanf("%d",&x);
if(x==1)
a++;//记录给定数列中1的个数
else
b++; //记录给定数列中-1的个数
}
for(i=0; i<m; i++)
{
scanf("%d%d",&x,&y);
if((y-x+1)%2==0&&(y-x+1)/2<=a&&(y-x+1)/2<=b)//l~r区间中的数正好有偶数个数
  //给定的数列中1，和-1的个数都不小于l~r区间中的数的个数的1/2
printf("1\n");
else
printf("0\n");
}
}
return 0;
}