lightoj1131

maksyuki 发表于 oj 分类，标签: 数值-矩阵快速幂取模

27 8月 2015

1131 - Just Two Functions

Find f_n % M and g_n % M. (% stands for the modulo operation.)Let f_n = a₁ * f_n-1 + b₁ * f_n-2 + c₁ * g_n-3g_n = a₂ * g_n-1 + b₂ * g_n-2 + c₂ * f_n-3

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

Each case starts with a blank line. Next line contains three integers a₁ b₁ c₁ (0 ≤ a₁, b₁, c₁ < 25000). Next line contains three integers a₂ b₂ c₂ (0 ≤ a₂, b₂, c₂ < 25000). Next line contains three integers f₀ f₁ f₂ (0 ≤ f₀, f₁, f₂ < 25000). Next line contains three integers g₀ g₁ g₂ (0 ≤ g₀, g₁, g₂ < 25000). The next line contains an integer M (1 ≤ M < 25000).

Next line contains an integer q (1 ≤ q ≤ 100) denoting the number of queries. Next line contains q space separated integers denoting n. Each of these integers is non-negative and less than 2³¹.

Output

For each case, print the case number in a line. Then for each query, you have to print one line containing f_n % M and g_n % M.

Sample Input

Output for Sample Input

21 1 00 0 00 1 1

0 0 0

20000

1 2 3 4 5 6 7 8 9 10

1 1 1

2 2 2

20000

2 4 6 8 10

Case 1:1 01 02 03 0

5 0

8 0

13 0

21 0

34 0

55 0

Case 2:

2 2

10 10

34 34

114 114

386 386

题目类型：矩阵快速幂取模

算法分析：由于两个递推式之间不是独立的，所以需要将两个递推式用一个矩阵乘幂表示！！！之后直接上模板即可

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

const int INF = 0x7FFFFFFF;
//const int MOD = 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 6;

long long n = 6;//表示方阵的尺寸，必须恰好为运算的大小

long long MOD;
//申请变量
struct Mat
{
    long long mat[maxn][maxn];
};


//重载Mat乘法运算
Mat operator * (Mat a, Mat b)
{
    Mat c;
    memset (c.mat, 0, sizeof (c.mat));
    for (long long k = 0; k < n; k++)
    {
        for (long long i = 0; i < n; i++)
        {
            if (a.mat[i][k] == 0) continue;

            for (long long j = 0; j < n; j++)
            {
                if (b.mat[k][j] == 0) continue;

                c.mat[i][j] = (c.mat[i][j] + ((a.mat[i][k] % MOD) * (b.mat[k][j] % MOD) % MOD)) % MOD;
            }
        }
    }
return c;
}

//重载Mat乘幂运算
Mat operator ^ (Mat a, long long k)
{
    Mat c;
    for (long long i = 0; i < n; i++)
        for(long long j = 0; j < n; j++)
        c.mat[i][j] = (i == j);

    while (k)
    {
        if(k & 1)
            c = c * a;

        a = a * a;
        k >>= 1;
    }

return c;
}

int main()
{
//	ifstream cin ("aaa.txt");
//	freopen ("aaa.txt", "r", stdin);

    int t, flag = 1;
    cin >> t;

    while(t--)
	{
        cout << "Case " << flag++ << ":" << endl;
        long long a1, b1, c1, a2, b2, c2, f0, f1, f2, g0, g1, g2;

        cin >> a1 >> b1 >> c1 >> a2 >> b2 >> c2 >> f0 >> f1 >> f2 >> g0 >> g1 >> g2 >> MOD;

		Mat aa, a0;
        aa.mat[0][0] = a1, aa.mat[0][1] = b1, aa.mat[0][2] = 0, aa.mat[0][3] = 0, aa.mat[0][4] = 0, aa.mat[0][5] = c1;
        aa.mat[1][0] = 1, aa.mat[1][1] = 0, aa.mat[1][2] = 0, aa.mat[1][3] = 0, aa.mat[1][4] = 0, aa.mat[1][5] = 0;
        aa.mat[2][0] = 0, aa.mat[2][1] = 1, aa.mat[2][2] = 0, aa.mat[2][3] = 0, aa.mat[2][4] = 0, aa.mat[2][5] = 0;
		aa.mat[3][0] = 0, aa.mat[3][1] = 0, aa.mat[3][2] = c2, aa.mat[3][3] = a2, aa.mat[3][4] = b2,  aa.mat[3][5] = 0;
		aa.mat[4][0] = 0, aa.mat[4][1] = 0, aa.mat[4][2] = 0, aa.mat[4][3] = 1, aa.mat[4][4] = 0, aa.mat[4][5] = 0;
		aa.mat[5][0] = 0, aa.mat[5][1] = 0, aa.mat[5][2] = 0, aa.mat[5][3] = 0, aa.mat[5][4] = 1, aa.mat[5][5] = 0;

		a0.mat[0][0] = f2, a0.mat[1][0] = f1, a0.mat[2][0] = f0, a0.mat[3][0] = g2, a0.mat[4][0] = g1, a0.mat[5][0] = g0;

        int q;
        cin >> q;

		long long tt;
		Mat res;
        for (int i = 1; i <= q; i++)
		{
			cin >> tt;
//			cout << tt << endl;

            if (tt == 0)
				cout << f0 % MOD << " " << g0 % MOD << endl;

			else if (tt == 1)
				cout << f1 % MOD << " " << g1 % MOD << endl;

			else if (tt == 2)
				cout << f2 % MOD << " " << g2 % MOD << endl;

			else
			{
				res = aa ^ (tt - 2);
				res = res * a0;
				cout << res.mat[0][0] % MOD << " " << res.mat[3][0] % MOD << endl;
			}
		}
	}
return 0;
}

100

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102

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109

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114

115

116

117

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142

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

const int INF = 0x7FFFFFFF;

//const int MOD = 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 6;

long long n = 6;//表示方阵的尺寸，必须恰好为运算的大小

long long MOD;

//申请变量

struct Mat

{

long long mat[maxn][maxn];

};

//重载Mat乘法运算

Mat operator * (Mat a, Mat b)

{

Mat c;

memset (c.mat, 0, sizeof (c.mat));

for (long long k = 0; k < n; k++)

{

for (long long i = 0; i < n; i++)

{

if (a.mat[i][k] == 0) continue;

for (long long j = 0; j < n; j++)

{

if (b.mat[k][j] == 0) continue;

c.mat[i][j] = (c.mat[i][j] + ((a.mat[i][k] % MOD) * (b.mat[k][j] % MOD) % MOD)) % MOD;

}

return c;

}

//重载Mat乘幂运算

Mat operator ^ (Mat a, long long k)

{

Mat c;

for (long long i = 0; i < n; i++)

for(long long j = 0; j < n; j++)

c.mat[i][j] = (i == j);

while (k)

{

if(k & 1)

c = c * a;

a = a * a;

k >>= 1;

}

return c;

}

int main()

{

// ifstream cin ("aaa.txt");

// freopen ("aaa.txt", "r", stdin);

int t, flag = 1;

cin >> t;

while(t--)

{

cout << "Case " << flag++ << ":" << endl;

long long a1, b1, c1, a2, b2, c2, f0, f1, f2, g0, g1, g2;

cin >> a1 >> b1 >> c1 >> a2 >> b2 >> c2 >> f0 >> f1 >> f2 >> g0 >> g1 >> g2 >> MOD;

Mat aa, a0;

aa.mat[0][0] = a1, aa.mat[0][1] = b1, aa.mat[0][2] = 0, aa.mat[0][3] = 0, aa.mat[0][4] = 0, aa.mat[0][5] = c1;

aa.mat[1][0] = 1, aa.mat[1][1] = 0, aa.mat[1][2] = 0, aa.mat[1][3] = 0, aa.mat[1][4] = 0, aa.mat[1][5] = 0;

aa.mat[2][0] = 0, aa.mat[2][1] = 1, aa.mat[2][2] = 0, aa.mat[2][3] = 0, aa.mat[2][4] = 0, aa.mat[2][5] = 0;

aa.mat[3][0] = 0, aa.mat[3][1] = 0, aa.mat[3][2] = c2, aa.mat[3][3] = a2, aa.mat[3][4] = b2, aa.mat[3][5] = 0;

aa.mat[4][0] = 0, aa.mat[4][1] = 0, aa.mat[4][2] = 0, aa.mat[4][3] = 1, aa.mat[4][4] = 0, aa.mat[4][5] = 0;

aa.mat[5][0] = 0, aa.mat[5][1] = 0, aa.mat[5][2] = 0, aa.mat[5][3] = 0, aa.mat[5][4] = 1, aa.mat[5][5] = 0;

a0.mat[0][0] = f2, a0.mat[1][0] = f1, a0.mat[2][0] = f0, a0.mat[3][0] = g2, a0.mat[4][0] = g1, a0.mat[5][0] = g0;

int q;

cin >> q;

long long tt;

Mat res;

for (int i = 1; i <= q; i++)

{

cin >> tt;

// cout << tt << endl;

if (tt == 0)

cout << f0 % MOD << " " << g0 % MOD << endl;

else if (tt == 1)

cout << f1 % MOD << " " << g1 % MOD << endl;

else if (tt == 2)

cout << f2 % MOD << " " << g2 % MOD << endl;

else

{

res = aa ^ (tt - 2);

res = res * a0;

cout << res.mat[0][0] % MOD << " " << res.mat[3][0] % MOD << endl;

}

return 0;

}

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lightoj1131

Input

Output

Sample Input

Output for Sample Input

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