E

题目链接：https://vjudge.net/problem/HDU-5950

思路： 构造矩阵，然后利用矩阵快速幂。

#include <bits/stdc++.h>
#include <time.h>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <utility>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <list>
using namespace std;
//cout<<setprecision(10)<<fixed;
#define eps 1e-6
#define PI acos(-1.0)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define mem(s,n) memset(s,n,sizeof s);
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rep2(i,a,b) for(int i=a;i>=b;i--)
#define ios {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}
typedef long long ll;
typedef unsigned long long ull;
const int maxn=1e6+5;
const ll Inf=0x7f7f7f7f7f7f7f7f;
const ll mod=2147493647;
//const int N=3e3+5;
bool isPowerOfTwo(int n) { return n > 0 && (n & (n - 1)) == 0; }//判断一个数是不是 2 的正整数次幂
int modPowerOfTwo(int x, int mod) { return x & (mod - 1); }//对 2 的非负整数次幂取模
int getBit(int a, int b) { return (a >> b) & 1; }// 获取 a 的第 b 位，最低位编号为 0
int Max(int a, int b) { return b & ((a - b) >> 31) | a & (~(a - b) >> 31); }// 如果 a>=b,(a-b)>>31 为 0，否则为 -1
int Min(int a, int b) { return a & ((a - b) >> 31) | b & (~(a - b) >> 31); }
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline int read()
{
    int X=0; bool flag=1; char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') flag=0; ch=getchar();}
    while(ch>='0'&&ch<='9') {X=(X<<1)+(X<<3)+ch-'0'; ch=getchar();}
    if(flag) return X;
    return ~(X-1);
}
inline void write(int X)
{
    if(X<0) {X=~(X-1); putchar('-');}
    if(X>9) write(X/10);
    putchar(X%10+'0');
}
/*
inline int write(int X)
{
    if(X<0) {putchar('-'); X=~(X-1);}
    int s[20],top=0;
    while(X) {s[++top]=X%10; X/=10;}
    if(!top) s[++top]=0;
    while(top) putchar(s[top--]+'0');
}
*/
void scan(__int128 &x)//输入
{
    x = 0;
    int f = 1;
    char ch;
    if((ch = getchar()) == '-') f = -f;
    else x = x*10 + ch-'0';
    while((ch = getchar()) >= '0' && ch <= '9')
        x = x*10 + ch-'0';
    x *= f;
}
void _print(__int128 x)
{
    if(x > 9) _print(x/10);
    putchar(x%10 + '0');
}
int Abs(int n) {
  return (n ^ (n >> 31)) - (n >> 31);
  /* n>>31 取得 n 的符号，若 n 为正数，n>>31 等于 0，若 n 为负数，n>>31 等于 -1
     若 n 为正数 n^0=n, 数不变，若 n 为负数有 n^(-1)
     需要计算 n 和 -1 的补码，然后进行异或运算，
     结果 n 变号并且为 n 的绝对值减 1，再减去 -1 就是绝对值 */
}
ll binpow(ll a, ll b) {
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = res * a%mod;
    a = a * a%mod;
    b >>= 1;
  }
  return res%mod;
}
void extend_gcd(ll a,ll b,ll &x,ll &y)
{
    if(b==0) {
        x=1,y=0;
        return;
    }
    extend_gcd(b,a%b,x,y);
    ll tmp=x;
    x=y;
    y=tmp-(a/b)*y;
}
ll mod_inverse(ll a,ll m)
{
    ll x,y;
    extend_gcd(a,m,x,y);
    return (m+x%m)%m;
}
ll eulor(ll x)
{
   ll cnt=x;
   ll ma=sqrt(x);
   for(int i=2;i<=ma;i++)
   {
    if(x%i==0) cnt=cnt/i*(i-1);
    while(x%i==0) x/=i;
   }
   if(x>1) cnt=cnt/x*(x-1);
   return cnt;
}
ll n,a,b;
typedef struct 
{
    ll mp[7][7];
    void init()
    {
        mem(mp,0);
        for(int i=0;i<7;i++)
           mp[i][i]=1;
    } 
}matrix;
matrix pp={
    1,1,0,0,0,0,0,
    2,0,0,0,0,0,0,
    1,0,1,0,0,0,0,
    4,0,4,1,0,0,0,
    6,0,6,3,1,0,0,
    4,0,4,3,2,1,0,
    1,0,1,1,1,1,1
};
matrix multi(matrix a,matrix b)
{   
    matrix res;
    for(int i=0;i<7;i++)
    {
        for(int j=0;j<7;j++)
        {
            res.mp[i][j]=0;
            for(int k=0;k<7;k++)
            {
                res.mp[i][j]=(res.mp[i][j]+(a.mp[i][k]*b.mp[k][j])%mod)%mod;
            }
        }
    }
    return res;
}
matrix fastm (matrix a,ll x)
{
    matrix res;
    res.init();
    while(x)
    {
        if(x&1) res=multi(res,a);
        x>>=1;
        a=multi(a,a);
    }
    return res;
}
int main()
{
    int t=read();
    while(t--)
    {
        scanf("%lld%lld%lld",&n,&a,&b);
        if(n==1) printf("%lld
",a);
        else if(n==2) printf("%lld
",b);
        else
        {
            matrix now=fastm(pp,n-2);
            ll num;
            num=(b*now.mp[0][0])%mod;
            num=(num+a*now.mp[1][0]%mod)%mod;
            num=(num+16*now.mp[2][0]%mod)%mod;
            num=(num+8*now.mp[3][0]%mod)%mod;
            num=(num+4*now.mp[4][0]%mod)%mod;
            num=(num+2*now.mp[5][0]%mod)%mod;
            num=(num+now.mp[6][0]%mod)%mod;
            printf("%lld
",num);
        }
    }
    return 0;
}