[Luogu] P1939 【模板】矩阵加速（数列）

题目描述

a[1]=a[2]=a[3]=1

a[x]=a[x-3]+a[x-1] (x>3)

求a数列的第n项对1000000007（10^9+7）取余的值。

题目解析

顺序:x.x，y.y，x.y

Code

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;const int MOD = 1000000007;struct Matrix {long long m[10][10];int sx,sy;
} str,pro;long long n;void init() {str.m[1][1] = str.m[1][2] = str.m[1][3] = 1;str.sx = 1;str.sy = 3;pro.m[1][1] = pro.m[1][2] = pro.m[2][2] = pro.m[2][3] = pro.m[3][1] = 0;pro.m[1][3] = pro.m[2][1] = pro.m[3][2] = pro.m[3][3] = 1;pro.sx = pro.sy = 3;return;
}inline Matrix mul(Matrix a,Matrix b) {Matrix res;memset(res.m,0,sizeof(res.m));for(int i = 1;i <= a.sx;i++) {for(int j = 1;j <= b.sy;j++) {for(int k = 1;k <= b.sx;k++) {res.m[i][j] += a.m[i][k] * b.m[k][j] % MOD;res.m[i][j] %= MOD;}}}res.sx = a.sx, res.sy = b.sy;return res;
}inline Matrix quick_pow(Matrix x,long long y) {Matrix res;res.sx = str.sx;res.sy = str.sy;for(int i = 1;i <= 5;i++) {for(int j = 1;j <= 5;j++) {res.m[i][j] = (i==j);}}while(y) {if(y & 1) res = mul(res,x);x = mul(x,x);y >>= 1;}return res;
}int main() {int T;scanf("%d",&T);while(T--) {scanf("%lld",&n);init();if(n <= 3) {puts("1");continue;}Matrix ans = mul(str,quick_pow(pro,n));printf("%lld\n",ans.m[1][3]);}return 0;
}