A.Eugeny and Array

              A. Eugeny and ArrayTime Limit:2000msCase Time Limit:2000msMemory Limit:262144KB64-bit integer IO format: %I64d      Java class name: (Any)SubmitStatusPID:
29452 Font Size:Eugeny has array a =  a 1, 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 integersl_{i,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 suma_{li + 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 ≤ 2·10^{5). The second line containsn integersa_{1, 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 31 -11 11 22 2

Output

010

Input

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

Output

01010

解題思路：本題為簡單題，只要搞懂題目，就能做出來。

給你一個數列，數組元素為1或-1，要你對數列進行重排，看能否滿足給定的標號l和r使得數列中l~r之間的數的和為0，若可以，輸出1，否則輸出0。

只要記錄數列中1，-1的個數然後根據給定的l和r判斷即可。若l~r區間中的數正好有偶數個數，而給定的數列中1，和-1的個數都不小於l~r區間中的數的個數的1/2，那麼，我們就可以重排序列使其滿足要求。如若這樣，輸出1，否則，輸出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++;   //記錄給定數列中1的個數            else                b++;    //記錄給定數列中-1的個數        }        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區間中的數正好有偶數個數              //給定的數列中1，和-1的個數都不小於l~r區間中的數的個數的1/2               printf("1\n");            else                printf("0\n");        }    }    return 0;}