- Recursive sequence
題意:給定起始的兩個數a,b,求第n個數%mod
思路:
第一眼看出來是矩陣快速冪,不過當時沒推出來(其實推了一半把自己給否定了)
正解是,根據二項式定理,對於f(n)=f(n-1)+2f(n-2)+n^4可以轉換爲
f(n)=f(n-1)+2f(n-2)+c(4,0)(n-1)^4+c(4,1)(n-1)^3+c(4,2)(n-1)^2+c(4,3)(n-1)+c(4,4) (#1)
那麼,對於下一項:
f(n+1)=f(n)+2f(n-1)+c(4,0)(n)^4+c(4,1)(n)^3+c(4,2)(n)^2+c(4,3)(n)+c(4,4) (#2)
可以發現#2的f(n),f(n-1),n^4,n^3,n^2,n,1都可以通過#1構造出來,那麼根據倍數關係構造相關的矩陣即可。
code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 2147493647ll;
struct Matirx{
int r, c;
ll mat[10][10];
};
Matirx Mul(Matirx a, Matirx b){
Matirx ret;
ret.r = a.r;
ret.c = b.c;
for(int i = 0; i < ret.r; i++){
for(int j = 0; j < ret.c; j++){
ret.mat[i][j] = 0;
for(int k = 0; k < a.c; k++){
ret.mat[i][j] += a.mat[i][k]*b.mat[k][j];
ret.mat[i][j] %= mod;
}
}
}
return ret;
}
Matirx Quick_pow(Matirx a, ll n){
Matirx ret;
ret.r = ret.c = 7;
memset(ret.mat, 0, sizeof(ret.mat));
for(int i = 0; i < 7; i++) ret.mat[i][i] = 1;
while(n){
if(n&1) ret = Mul(ret, a);
a = Mul(a, a);
n >>= 1;
}
return ret;
}
void build(Matirx& t){
t.r = t.c = 7;
t.mat[0][0] = 1, t.mat[0][1] = 4, t.mat[0][2] = 6, t.mat[0][3] = 4, t.mat[0][4] = 1, t.mat[0][5] = 0, t.mat[0][6] = 0;//1
t.mat[1][0] = 0, t.mat[1][1] = 1, t.mat[1][2] = 3, t.mat[1][3] = 3, t.mat[1][4] = 1, t.mat[1][5] = 0, t.mat[1][6] = 0;//2
t.mat[2][0] = 0, t.mat[2][1] = 0, t.mat[2][2] = 1, t.mat[2][3] = 2, t.mat[2][4] = 1, t.mat[2][5] = 0, t.mat[2][6] = 0;//2
t.mat[3][0] = 0, t.mat[3][1] = 0, t.mat[3][2] = 0, t.mat[3][3] = 1, t.mat[3][4] = 1, t.mat[3][5] = 0, t.mat[3][6] = 0;//4
t.mat[4][0] = 0, t.mat[4][1] = 0, t.mat[4][2] = 0, t.mat[4][3] = 0, t.mat[4][4] = 1, t.mat[4][5] = 0, t.mat[4][6] = 0;//5
t.mat[5][0] = 0, t.mat[5][1] = 0, t.mat[5][2] = 0, t.mat[5][3] = 0, t.mat[5][4] = 0, t.mat[5][5] = 0, t.mat[5][6] = 1;//6
t.mat[6][0] = 1, t.mat[6][1] = 4, t.mat[6][2] = 6, t.mat[6][3] = 4, t.mat[6][4] = 1, t.mat[6][5] = 2, t.mat[6][6] = 1;//7
}
int main(){
int T;
scanf("%d", &T);
ll N, a, b;
while(T--){
scanf("%I64d%I64d%I64d", &N, &a, &b);
if(N == 1) printf("%I64d\n", a%mod);
else if(N == 2) printf("%I64d\n", b%mod);
else{
Matirx temp;
build(temp);
Matirx ret = Quick_pow(temp, N-2);
temp.r = 7;
temp.c = 1;
temp.mat[0][0] = 16,temp.mat[1][0] = 8,temp.mat[2][0] = 4,temp.mat[3][0] = 2,temp.mat[4][0] = 1,temp.mat[5][0] = a%mod, temp.mat[6][0] = b%mod;
ret = Mul(ret, temp);
printf("%I64d\n", ret.mat[6][0]);
}
}
return 0;
}