矩阵算法的基本题型

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1575

很水的模板题：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define maxn 101#define MOD 9973struct Mat{    int a[maxn][maxn];    Mat() {memset(a, 0, sizeof(a));}    Mat(int i)    {        memset(a, 0, sizeof(a));        for (int i=0; i<maxn; ++i)            a[i][i] = 1;    }}E(1);Mat operator*(Mat a, Mat b){    Mat c;    for (int i=0; i<maxn; ++i)        for (int j=0; j<maxn; ++j)        {            c.a[i][j] = 0;            for (int k=0; k<maxn; ++k)            {                c.a[i][j] += a.a[i][k]*b.a[k][j];                if (c.a[i][j]>=MOD)                    c.a[i][j] %= MOD;            }        }    return c;}Mat operator^(Mat a, int n){    Mat ans = E, q = a;    for (; n; n>>=1)    {        if (n&1)            ans = ans*q;        q = q*q;    }    return ans;}int main(){    int ncase, n, k, ans;    Mat a;    cin >> ncase;    while (ncase--)    {        cin >> n >> k;        for (int i=0; i<n; ++i)            for (int j=0; j<n; ++j)                cin >> a.a[i][j];        a = a^k;        ans = 0;        for (int i=0; i<n; ++i)            ans += a.a[i][i];        ans%=MOD;        cout << ans << endl;    }    return 0;}

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1757

构造一个

|f(10) |    |a0 a1 a2 ...a8 a9| |f(9)|
| f(9)  |    | 1 0 0 ..  .     0 0    | |f(8)|
| .....   | = | .. ......................... |  | ..  |
| f(2)  |    | 0 0 0 ...           0 0|  |f(1)|
| f(1)  |    | 0 0 0 ...          1 0|  |f(0)|

#include <iostream>#include <cstdio>using namespace std;#define N 10int MOD = 9973;struct Mat{    int mat[N][N];    Mat()    {        memset(mat, 0, sizeof(mat));    }}E;void get_E(Mat &a){    for (int i=0; i<N; ++i)        a.mat[i][i] = 1;}Mat operator*(Mat a, Mat b){    Mat c;    for (int i=0; i<N; ++i)        for (int j=0; j<N; ++j)        {            c.mat[i][j] = 0;            for (int k=0; k<N; ++k)                c.mat[i][j] += (a.mat[i][k]*b.mat[k][j])%MOD;            c.mat[i][j] %= MOD;        }    return c;}Mat operator^(Mat a, int n){    Mat ans = E, p = a;    for ( ; n; n>>=1)    {        if (n&1)            ans = ans*p;        p = p*p;    }    return ans;}int main(){    get_E(E);    int k, m, a[N];    while (cin >> k >> m)    {        for (int i=0; i<N; ++i)            cin >> a[i];        MOD = m;        if (k<10)        {            cout << k%m << endl;            continue;        }        Mat p;        for (int i=0; i<N; ++i)            p.mat[0][i] = a[i];        for (int i=1; i<N; ++i)            p.mat[i][i-1] = 1;        k = k-9;        p = p^k;        int ans = 0;        for (int i=0; i<N; i++)            ans += (p.mat[0][i]*(N-1-i))%MOD;        cout << ans%MOD << endl;    }    return 0;}

题目链接：http://acm.fzu.edu.cn/problem.php?pid=1683

构造

3 2 7 0    f(n-1)        f(n)

1 0 0 0 *  f(n-2)  =   f(n-1)

0 1 0 0    f(n-3)       f(n-2)

3 2 7 1    sum         sum'

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define maxn 4#define MOD 2009struct Mat{    int a[maxn][maxn];    Mat() {memset(a, 0, sizeof(a));}    Mat(int i)    {        memset(a, 0, sizeof(a));        for (int i=0; i<maxn; ++i)            a[i][i] = 1;    }}E(1);Mat operator*(Mat a, Mat b){    Mat c;    for (int i=0; i<maxn; ++i)        for (int j=0; j<maxn; ++j)        {            c.a[i][j] = 0;            for (int k=0; k<maxn; ++k)            {                c.a[i][j] += a.a[i][k]*b.a[k][j];                if (c.a[i][j]>=MOD)                    c.a[i][j] %= MOD;            }        }    return c;}Mat operator^(Mat a, int n){    Mat ans = E, q = a;    for (; n; n>>=1)    {        if (n&1)            ans = ans*q;        q = q*q;    }    return ans;}int main(){    int ncase, n, f[4] = {5, 3, 1, 9};    scanf("%d", &ncase);    for (int k=1; k<=ncase; ++k)    {        scanf("%d", &n);        int ans = 0;        if (n<3)        {//            for (int i=0; i<=n; ++i)//                ans += f[i];            printf("Case %d: %d\n", k, f[2-n]);        }        else        {            Mat p;            p.a[0][0] = p.a[3][0] =  3;            p.a[0][1] = p.a[3][1] =  2;            p.a[0][2] = p.a[3][2] =  7;            p.a[1][0] = p.a[3][3] =  1;            p.a[2][1] = 1;            n -= 2;            p = p^n;//            for (int i=0; i<maxn; i++)//            {//                for (int j=0; j<maxn; j++)//                    cout << p.a[i][j] << " ";//                cout << endl;//            }            ans += (p.a[3][0]*5)%MOD;            ans += (p.a[3][1]*3)%MOD;            ans += (p.a[3][2]*1)%MOD;            ans += (p.a[3][3]*9)%MOD;            printf("Case %d: %d\n", k, ans%MOD);        }    }    return 0;}