http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4108

题意：求斐波那契数列区间[l，r]之和的奇偶性

题解： 规律

1、斐波那契数列奇偶性为 1 1 0 1 1 0 1 1 0 ......；

2、在区间，如果时为奇数，时为偶数；

3、询问一个数的余数=各位数字之和；

4、奇数-奇数=奇数，偶数-偶数=偶数，奇数-偶数=奇数，偶数-奇数=奇数；

5、询问和的奇偶性，不相同则1，相同则0；

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
char a[N],b[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    scanf("%d",&t);
    while(t--){
        //scanf("%d",&n);
        cin>>a>>b;
        sum=0;
        ans=0;
        for(int i=0;a[i]!='\0';i++){
            sum+=a[i]-'0';
        }
        for(int i=0;b[i]!='\0';i++){
            ans+=b[i]-'0';
        }
        sum%=3;
        sum=(sum+2)%3;
        ans%=3;
        sum=sum==1;
        ans=ans==1;
        cout<<(ans!=sum)<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}