http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1002&cid=843

题解：斐波那契数列+前缀和

/*
*@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
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=40000+10;
const int M=100000+10;
const int MOD=192600817;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q;
int ans,cnt,flag,temp,sum;
ll f[N];
ll ff[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);

    ff[1]=f[1]=f[0]=1;
    ff[2]=2;
    for(int i=2;i<=40000+3;i++){
        f[i]=(f[i-1]+f[i-2])%MOD;
        ff[i+1]=((f[i]*f[i])%MOD+ff[i])%MOD;
    }
    while(~scanf("%d",&t)){
        while(t--){
            int a,b,c,d;
            scanf("%d%d%d%d",&a,&b,&c,&d);
            int x=a*4+b;
            int y=c*4+d;
            //cout<<x<<" "<<y<<endl;
            printf("%lld\n",(ff[max(y,x)+1]-ff[min(x,y)]+MOD)%MOD);
        }
    }


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