Reading comprehension

這個不難找出遞推式$f[n]=f[n?1]+2f[n?2]+1$。

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f; const double eps = 1e-7;ll n, mod;struct matrix {ll a[3][3]; };matrix operator * (matrix a, matrix b) {matrix ans;for(int i = 0; i < 3; i++) {for(int j = 0; j < 3; j++) {ans.a[i][j] = 0;for(int k = 0; k < 3; k++) {ans.a[i][j] = (ans.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % mod) % mod;}}}return ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);while(scanf("%lld %lld", &n, &mod) != EOF && (n + mod)) {if(n <= 2) {if(n == 1) printf("%d\n", 1 % mod);else printf("%d\n", 2 % mod);continue;}matrix ans = {2, 1, 1,0, 0, 0,0, 0, 0};matrix a = {1, 1, 0,2, 0, 0, 1, 0, 1};n -= 2;while(n) {if(n & 1) ans = ans * a;a = a * a;n >>= 1;}printf("%lld\n", ans.a[0][0]);}return 0; }

233 Matrix

這里直接寫出構造的矩陣了。
$$A=?\begin{array}{ccccc}{a}_{00}& a_{10}& a_{20}\dots a_{(n?1)0}& a_{n0}& 1\\ 0& \dots & & & \\ 0& \dots & & & \end{array}?$$
整體來說就是一個$(n+2)\times (n+2)$的矩陣吧。
$$B=?\begin{array}{ccccc}10& 10& 10\dots 10& 10& 0\\ 0& 1& 1\dots 1& 1& 0\\ 0& 0& 1\dots 1& 1& 0\\ \dots & & & & \\ 3& 3& 3\dots 3& 3& 1\end{array}?$$
同樣的也是一個$(n+2)\times (n+2)$的矩陣。

我們個$a_{00}$初始化為$23$，這個時候$A\times B$發現$A$中的元素會變成第二列的，所以這里就形成了一個遞推式了，只要跑一跑矩陣快速冪即可。

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f; const double eps = 1e-7;const int mod = 10000007;int n, m;struct matrix {ll a[15][15];void init() {memset(a, 0, sizeof a);} };matrix operator * (matrix a, matrix b) {matrix ans;for(int i = 0; i < n; i++) {for(int j = 0; j < n; j++) {ans.a[i][j] = 0;for(int k = 0; k < n; k++) {ans.a[i][j] = (ans.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % mod) % mod;}}}return ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);while(scanf("%d %d", &n, &m) != EOF) {matrix ans, a;ans.init(), a.init();for(int i = 1; i <= n; i++) scanf("%lld", &ans.a[0][i]);ans.a[0][0] = 23, ans.a[0][n + 1] = 1;for(int i = 0; i <= n; i++) {a.a[0][i] = 10;a.a[n + 1][i] = 3;}a.a[n + 1][n + 1] = 1;for(int i = 1; i <= n; i++) {for(int j = i; j <= n; j++) {a.a[i][j] = 1;}}n += 2;while(m) {if(m & 1) ans = ans * a;a = a * a;m >>= 1;}printf("%lld\n", ans.a[0][n - 2]);}return 0; }

B. Jzzhu and Sequences

裸題$f_n=f_{n?1}?f_{n?2}$。

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f; const double eps = 1e-7;const int mod = 1e9 + 7;int n = 2, m;struct matrix {ll a[2][2];void init() {memset(a, 0, sizeof a);} };matrix operator * (matrix a, matrix b) {matrix ans;for(int i = 0; i < n; i++) {for(int j = 0; j < n; j++) {ans.a[i][j] = 0;for(int k = 0; k < n; k++) {ans.a[i][j] = (ans.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % mod) % mod;}}}return ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);ll x, y, n;scanf("%lld %lld %lld", &x, &y, &n);if(n <= 2) {if(n == 1) printf("%d\n", (x % mod + mod) % mod);else printf("%d\n", (y % mod + mod) % mod);return 0;}matrix ans = {y, x, 0, 0};matrix a = {1, 1,-1, 0};n -= 2;while(n) {if(n & 1) ans = ans * a;a = a * a;n >>= 1;}printf("%lld", (ans.a[0][0] % mod + mod) % mod);return 0; }

M斐波那契數列

這是個乘法的遞推式，顯然乘法可以變成指數相加，所以還是一個矩陣快速冪，統計最后一項有多少個$A_0,A_1$即可。

當然矩陣中的取模得用費馬小定理。

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f; const double eps = 1e-7;const int mod = 1000000006;int n = 4, m;struct matrix {ll a[4][4];void init() {memset(a, 0, sizeof a);} };matrix operator * (matrix a, matrix b) {matrix ans;for(int i = 0; i < n; i++) {for(int j = 0; j < n; j++) {ans.a[i][j] = 0;for(int k = 0; k < n; k++) {ans.a[i][j] = (ans.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % mod) % mod;}}}return ans; }ll quick_pow(ll a, int n) {ll ans = 1;const int mod = 1e9 + 7;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);ll x, y, n;while(scanf("%lld %lld %lld", &x, &y, &n) != EOF) {const int mod = 1e9 + 7;if(n <= 1) {if(n == 0) {printf("%lld\n", x % mod);}else {printf("%lld\n", y % mod);}continue;}matrix ans = {1, 0, 0, 1,0, 0, 0, 0,0, 0, 0, 0,0, 0, 0, 0};matrix a = {1, 0, 1, 0,0, 1, 0, 1,1, 0, 0, 0,0, 1, 0, 0};n -= 1;while(n) {if(n & 1) ans = ans * a;a = a * a;n >>= 1;}ll res = 1ll * quick_pow(x, ans.a[0][1]) * quick_pow(y, ans.a[0][0]) % mod;printf("%lld\n", res);}return 0; }

So Easy!

共軛矩陣構造的經典題了。

假設$A_n=(a+\sqrt{b}{)}^n,B_n=(a?\sqrt{b}{)}^n$，一定有$A_n+B_n$是個有理數

并且有$B_n<1$，所以有$?S_n?=A_n+B_n$

$2a{S}_n=((a+\sqrt{b}{)}^n+(a?\sqrt{b}{)}^n)((a+\sqrt{b})+(a?\sqrt{b}))=S_{n+1}+(a^2?b)S_{n?1}$

得到$S_n=2a{S}_{n?1}+(b?a^2)S_{n?2}$，于是遞推式就得到了進行矩陣快速冪即可。

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f; const double eps = 1e-7;// const int mod = 1000000006;ll n = 2, mod;struct matrix {ll a[2][2];void init() {memset(a, 0, sizeof a);} };matrix operator * (matrix a, matrix b) {matrix ans;for(int i = 0; i < n; i++) {for(int j = 0; j < n; j++) {ans.a[i][j] = 0;for(int k = 0; k < n; k++) {ans.a[i][j] = (ans.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % mod) % mod;}}}return ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);ll a, b, n;while(scanf("%lld %lld %lld %lld", &a, &b, &n, &mod) != EOF) {ll a1 = 2ll * a % mod, a2 = 2ll * (a * a % mod + b) % mod;if(n <= 2) {if(n == 1) printf("%lld\n", a1);else printf("%lld\n", a2);continue;}n -= 2;matrix ans = {a2, a1, 0, 0};matrix fat = {2ll * a % mod, 1,((b - a * a % mod) % mod + mod) % mod, 0};while(n) {if(n & 1) ans = ans * fat;fat = fat * fat;n >>= 1;}printf("%lld\n", ans.a[0][0]);}return 0; }

Problem of Precision

這道題目是向下取整，跟上一道題目一樣構造，唯一的不同就是$S_n=A_n+B_n?1$，所以我們還是按照$S_n$遞推，最后在答案上減一即可。

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f; const double eps = 1e-7;const int mod = 1024;int n = 2;struct matrix {ll a[2][2];void init() {memset(a, 0, sizeof a);} };matrix operator * (matrix a, matrix b) {matrix ans;for(int i = 0; i < n; i++) {for(int j = 0; j < n; j++) {ans.a[i][j] = 0;for(int k = 0; k < n; k++) {ans.a[i][j] = (ans.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % mod) % mod;}}}return ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T;scanf("%d", &T);while(T--) {ll n;scanf("%lld", &n);ll a1 = 10, a2 = 98;if(n <= 2) {if(n == 1) printf("%lld\n", (a1 - 1) % mod);else printf("%lld\n", (a2 - 1) % mod);continue;}n -= 2;matrix ans = {a2, a1,0, 0};matrix fat = {10, 1,1023, 0};while(n) {if(n & 1) ans = ans * fat;fat = fat * fat;n >>= 1;}printf("%lld\n", (ans.a[0][0] - 1 + mod) % mod);}return 0; }

C. Partial Sums

由于遞推矩陣的特殊關系，所以可以轉化為$1\times n$的矩陣遞推，最后達到$n^{2}log(n)$的復雜度。

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f; const double eps = 1e-7;const int N = 2e3 + 10, mod = 1e9 + 7;int n, k;struct matrix {ll a[N]; };matrix operator * (matrix a, matrix b) {matrix ans;for(int i = 1; i <= n; i++) {ans.a[i] = 0;for(int j = 1; j <= i; j++) {ans.a[i] = (ans.a[i] + a.a[j] * b.a[i - j + 1] % mod) % mod;}}return ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d %d", &n, &k);matrix ans, a;for(int i = 1; i <= n; i++) {scanf("%d", &ans.a[i]);a.a[i] = 1;}while(k) {if(k & 1) ans = ans * a;a = a * a;k >>= 1;}for(int i = 1; i <= n; i++) {printf("%lld%c", ans.a[i], i == n ? '\n' : ' ');}return 0; }

Fast Matrix Calculation

$A{B}^{n^2}=A(BA{)}^{n^2?1}B$,轉化為$k\times k$的矩陣快速冪，這樣就可以保證不超時了。

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f; const double eps = 1e-7;const int N = 1e3 + 10, mod = 6;int n, m;struct matrix {ll a[6][6];void init() {memset(a, 0, sizeof a);} }c;struct Matrix {ll a[N][N]; }a, b, Ans1, Ans2;matrix operator * (matrix a, matrix b) {matrix ans;for(int i = 0; i < m; i++) {for(int j = 0; j < m; j++) {ans.a[i][j] = 0;for(int k = 0; k < m; k++) {ans.a[i][j] = (ans.a[i][j] + a.a[i][k] * b.a[k][j] % mod) % mod;}}}return ans; }matrix mult() {matrix ans;for(int i = 0; i < m; i++) {for(int j = 0; j < m; j++) {ans.a[i][j] = 0;for(int k = 0; k < n; k++) {ans.a[i][j] = (ans.a[i][j] + b.a[i][k] * a.a[k][j] % mod) % mod;}}}return ans; }int main() {// freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);while(scanf("%d %d", &n, &m) && (n + m)) {for(int i = 0; i < n; i++) {for(int j = 0; j < m; j++) {scanf("%lld", &a.a[i][j]);}}for(int i = 0; i < m; i++) {for(int j = 0; j < n; j++) {scanf("%lld", &b.a[i][j]);}}c = mult();matrix ans;ans.init();for(int i = 0; i < m; i++) ans.a[i][i] = 1;int n1 = n * n - 1;while(n1) {if(n1 & 1) ans = ans * c;c = c * c;n1 >>= 1;}for(int i = 0; i < n; i++) {for(int j = 0; j < m; j++) {Ans1.a[i][j] = 0;for(int k = 0; k < m; k++) {Ans1.a[i][j] = (Ans1.a[i][j] + a.a[i][k] * ans.a[k][j] % mod) % mod;}}}for(int i = 0; i < n; i++) {for(int j = 0; j < n; j++) {Ans2.a[i][j] = 0;for(int k = 0; k < m; k++) {Ans2.a[i][j] = (Ans2.a[i][j] + Ans1.a[i][k] * b.a[k][j] % mod) % mod;}}}int res = 0;for(int i = 0; i < n; i++) {for(int j = 0; j < n; j++) {res += Ans2.a[i][j];}}printf("%d\n", res);}return 0; }

總結

以上是生活随笔為你收集整理的矩阵快速幂各类题型总结（一般，共轭，1 * n， 矩阵简化）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。