E - Recursive sequence HDU - 5950 （矩阵快速幂）

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。

  1 #include <bits/stdc++.h>
  2 #include <time.h>
  3 #include <set>
  4 #include <map>
  5 #include <stack>
  6 #include <cmath>
  7 #include <queue>
  8 #include <cstdio>
  9 #include <string>
 10 #include <vector>
 11 #include <cstring>
 12 #include <utility>
 13 #include <cstring>
 14 #include <iostream>
 15 #include <algorithm>
 16 #include <list>
 17 using namespace std;
 18 //cout<<setprecision(10)<<fixed;
 19 #define eps 1e-6
 20 #define PI acos(-1.0)
 21 #define lowbit(x) ((x)&(-x))
 22 #define zero(x) (((x)>0?(x):-(x))<eps)
 23 #define mem(s,n) memset(s,n,sizeof s);
 24 #define rep(i,a,b) for(int i=a;i<=b;i++)
 25 #define rep2(i,a,b) for(int i=a;i>=b;i--)
 26 #define ios {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}
 27 typedef long long ll;
 28 typedef unsigned long long ull;
 29 const int maxn=1e6+5;
 30 const ll Inf=0x7f7f7f7f7f7f7f7f;
 31 const ll mod=2147493647;
 32 //const int N=3e3+5;
 33 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
 34 int modPowerOfTwo(int x, int mod) { return x & (mod - 1); }//对 2 的非负整数次幂取模
 35 int getBit(int a, int b) { return (a >> b) & 1; }// 获取 a 的第 b 位，最低位编号为 0
 36 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
 37 int Min(int a, int b) { return a & ((a - b) >> 31) | b & (~(a - b) >> 31); }
 38 ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
 39 ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
 40 inline int read()
 41 {
 42     int X=0; bool flag=1; char ch=getchar();
 43     while(ch<'0'||ch>'9') {if(ch=='-') flag=0; ch=getchar();}
 44     while(ch>='0'&&ch<='9') {X=(X<<1)+(X<<3)+ch-'0'; ch=getchar();}
 45     if(flag) return X;
 46     return ~(X-1);
 47 }
 48 inline void write(int X)
 49 {
 50     if(X<0) {X=~(X-1); putchar('-');}
 51     if(X>9) write(X/10);
 52     putchar(X%10+'0');
 53 }
 54 /*
 55 inline int write(int X)
 56 {
 57     if(X<0) {putchar('-'); X=~(X-1);}
 58     int s[20],top=0;
 59     while(X) {s[++top]=X%10; X/=10;}
 60     if(!top) s[++top]=0;
 61     while(top) putchar(s[top--]+'0');
 62 }
 63 */
 64 void scan(__int128 &x)//输入
 65 {
 66     x = 0;
 67     int f = 1;
 68     char ch;
 69     if((ch = getchar()) == '-') f = -f;
 70     else x = x*10 + ch-'0';
 71     while((ch = getchar()) >= '0' && ch <= '9')
 72         x = x*10 + ch-'0';
 73     x *= f;
 74 }
 75 void _print(__int128 x)
 76 {
 77     if(x > 9) _print(x/10);
 78     putchar(x%10 + '0');
 79 }
 80 int Abs(int n) {
 81   return (n ^ (n >> 31)) - (n >> 31);
 82   /* 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
 83      若 n 为正数 n^0=n, 数不变，若 n 为负数有 n^(-1)
 84      需要计算 n 和 -1 的补码，然后进行异或运算，
 85      结果 n 变号并且为 n 的绝对值减 1，再减去 -1 就是绝对值 */
 86 }
 87 ll binpow(ll a, ll b) {
 88   ll res = 1;
 89   while (b > 0) {
 90     if (b & 1) res = res * a%mod;
 91     a = a * a%mod;
 92     b >>= 1;
 93   }
 94   return res%mod;
 95 }
 96 void extend_gcd(ll a,ll b,ll &x,ll &y)
 97 {
 98     if(b==0) {
 99         x=1,y=0;
100         return;
101     }
102     extend_gcd(b,a%b,x,y);
103     ll tmp=x;
104     x=y;
105     y=tmp-(a/b)*y;
106 }
107 ll mod_inverse(ll a,ll m)
108 {
109     ll x,y;
110     extend_gcd(a,m,x,y);
111     return (m+x%m)%m;
112 }
113 ll eulor(ll x)
114 {
115    ll cnt=x;
116    ll ma=sqrt(x);
117    for(int i=2;i<=ma;i++)
118    {
119     if(x%i==0) cnt=cnt/i*(i-1);
120     while(x%i==0) x/=i;
121    }
122    if(x>1) cnt=cnt/x*(x-1);
123    return cnt;
124 }
125 ll n,a,b;
126 typedef struct
127 {
128     ll mp[7][7];
129     void init()
130     {
131         mem(mp,0);
132         for(int i=0;i<7;i++)
133            mp[i][i]=1;
134     }
135 }matrix;
136 matrix pp={
137     1,1,0,0,0,0,0,
138     2,0,0,0,0,0,0,
139     1,0,1,0,0,0,0,
140     4,0,4,1,0,0,0,
141     6,0,6,3,1,0,0,
142     4,0,4,3,2,1,0,
143     1,0,1,1,1,1,1
144 };
145 matrix multi(matrix a,matrix b)
146 {
147     matrix res;
148     for(int i=0;i<7;i++)
149     {
150         for(int j=0;j<7;j++)
151         {
152             res.mp[i][j]=0;
153             for(int k=0;k<7;k++)
154             {
155                 res.mp[i][j]=(res.mp[i][j]+(a.mp[i][k]*b.mp[k][j])%mod)%mod;
156             }
157         }
158     }
159     return res;
160 }
161 matrix fastm (matrix a,ll x)
162 {
163     matrix res;
164     res.init();
165     while(x)
166     {
167         if(x&1) res=multi(res,a);
168         x>>=1;
169         a=multi(a,a);
170     }
171     return res;
172 }
173 int main()
174 {
175     int t=read();
176     while(t--)
177     {
178         scanf("%lld%lld%lld",&n,&a,&b);
179         if(n==1) printf("%lld\n",a);
180         else if(n==2) printf("%lld\n",b);
181         else
182         {
183             matrix now=fastm(pp,n-2);
184             ll num;
185             num=(b*now.mp[0][0])%mod;
186             num=(num+a*now.mp[1][0]%mod)%mod;
187             num=(num+16*now.mp[2][0]%mod)%mod;
188             num=(num+8*now.mp[3][0]%mod)%mod;
189             num=(num+4*now.mp[4][0]%mod)%mod;
190             num=(num+2*now.mp[5][0]%mod)%mod;
191             num=(num+now.mp[6][0]%mod)%mod;
192             printf("%lld\n",num);
193         }
194     }
195     return 0;
196 }