poj 3233 Matrix Power Series

Matrix Power Series

思路

题意比较简单，就是要求S(n)=∑i=1nAiS(n) = \sum _{i = 1} ^{n} A^ {i}S(n)=∑i=1nAi，显然有S(n)=S(n−1)∗A+AS(n) = S(n - 1) * A + AS(n)=S(n−1)∗A+A，看到这里，那就简单了，递推式，加矩阵，矩阵快速幂无疑了嘛，所以我们开始构造矩阵。

显然有如下矩阵，EEE是单位矩阵，AAA是输入矩阵，OOO是零矩阵。

[EEOA]∗[OOAO]\begin{bmatrix} E & E \\ O & A\end{bmatrix} * \begin{bmatrix} O & O\\ A & O \end{bmatrix}[EOEA]∗[OAOO]

通过这个矩阵的递推，我们就可以通过快速幂，达到快速求解的目的。

我严重怀疑这道题目数据有问题，longlonglong\ longlong long就wawawa，然后intintint就过了？？？

AC代码

/*Author : lifehappy
*/
// #pragma GCC optimize(2)
// #pragma GCC optimize(3)
// #include <bits/stdc++.h>#include <cstdio>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
#include <cmath>#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 70;int n, k, mod;struct matrix {int a[N][N];matrix operator * (const matrix & t) const {matrix temp;for(int i = 1; i <= 2 * n; i++) {for(int j = 1; j <= 2 * n; j++) {temp.a[i][j] = 0;for(int k = 1; k <= 2 * n; k++) {temp.a[i][j] = (temp.a[i][j] + a[i][k] * t.a[k][j]) % mod;}}}return temp;}
}E, A, O;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), k = read(), mod = read();matrix fat, ans;for(int i = 1; i <= 2 * n; i++) {//先置零，for(int j = 1; j <= 2 * n; j++) {fat.a[i][j] = ans.a[i][j] = 0;}}for(int i = 1; i <= n; i++) {//读入的时候置A加上置E矩阵。for(int j = 1; j <= n; j++) {fat.a[i + n][j + n] = ans.a[i + n][j] = read();}fat.a[i][i] = fat.a[i][i + n] = 1;}while(k) {if(k & 1) ans = fat * ans;fat = fat * fat;k >>= 1;}for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {printf("%d%c", ans.a[i][j], j == n ? '\n' : ' ');}}return 0;
}

调不出来的代码

写了一手逼格高一点的举证套矩阵的重载操作符的写法，可是太菜了，调不出来

/*Author : lifehappy
*/
// #pragma GCC optimize(2)
// #pragma GCC optimize(3)
// #include <bits/stdc++.h>#include <cstdio>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
#include <cmath>#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 70;int n, k, mod;struct matrix {int a[N][N];matrix operator * (const matrix & t) const {matrix temp;for(int i = 1; i <= 2 * n; i++) {for(int j = 1; j <= 2 * n; j++) {temp.a[i][j] = 0;for(int k = 1; k <= 2 * n; k++) {temp.a[i][j] = (temp.a[i][j] + a[i][k] * t.a[k][j]) % mod;}}}return temp;}matrix operator + (const matrix & t) const {matrix temp;for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {temp.a[i][j] = (a[i][j] + t.a[i][j]) % mod;}}return temp;}
}E, A, O;struct Matrix {matrix a[3][3];Matrix operator * (const Matrix & t) const {Matrix temp;for(int i = 1; i <= 2; i++) {for(int j = 1; j <= 2; j++) {temp.a[i][j] = O;for(int k = 1; k <= 2; k++) {temp.a[i][j] = (a[i][k] * t.a[k][j]) + temp.a[i][j];}}}}
};int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), k = read(), mod = read();for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {A.a[i][j] = read();E.a[i][j] = O.a[i][j] = 0;}E.a[i][i] = 1;}Matrix fat, ans;fat.a[1][1] = E, fat.a[1][2] = E, fat.a[2][1] = O, fat.a[2][2] = A;ans.a[1][1] = O, ans.a[1][2] = O, ans.a[2][1] = A, ans.a[2][2] = O;while(k) {if(k & 1) ans = ans * fat;fat = fat * fat;k >>= 1;}for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {printf("%d%c", ans.a[1][1].a[i][j], j == n ? '\n' : ' ');}}return 0;
}

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poj 3233 Matrix Power Series

Matrix Power Series

思路

AC代码

调不出来的代码

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