CF:Eugeny and Array

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 mlines 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 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

解题报告：
大概的意思是先给出一个串a（1or-1),然后给出m行l,r两个数，然后重新对a进行排列,如果al+..........+ar=0；输出1否则输出0；
思路，如果r-l+1是奇数，那么数列和肯定不能为0，如果是偶数的话判断一下-1或者1的数量是否够用。
参考代码：
#include <stdio.h>

#include <string.h>

#include <iostream>

#include <cmath>

using namespace std;

int main(){

   // freopen("in.txt","r",stdin);

   int n,m,i,j,l,r,t,n1,m1,count,m_min;

   while(cin>>n>>m)

   {

     l=0;

     r=0;

     n1=0;

     m1=0;

     for(i=0;i<n;i++)

     {

         cin>>t;

         if(t==-1)

        n1++;

        else

        m1++;

     }

     for(i=0;i<m;i++)

     {

         cin>>l>>r;

         count=r-l+1;

       if(count%2!=0)

       {

     cout<<"0"<<endl;

       continue;

       }

       m_min=min(n1,m1);

       if(count/2<=m_min)

       cout<<"1"<<endl;

       else

       cout<<"0"<<endl;

     }

   }

    return 0;

}