LightOJ1006---Hex-a-bonacci(矩阵快速幂)

Given a code (not optimized), and necessary inputs, you have to find the output of the code for the inputs. The code is as follows:

int a, b, c, d, e, f;
int fn( int n ) {
    if( n == 0 ) return a;
    if( n == 1 ) return b;
    if( n == 2 ) return c;
    if( n == 3 ) return d;
    if( n == 4 ) return e;
    if( n == 5 ) return f;
    return( fn(n-1) + fn(n-2) + fn(n-3) + fn(n-4) + fn(n-5) + fn(n-6) );
}
int main() {
    int n, caseno = 0, cases;
    scanf("%d", &cases);
    while( cases-- ) {
        scanf("%d %d %d %d %d %d %d", &a, &b, &c, &d, &e, &f, &n);
        printf("Case %d: %d\n", ++caseno, fn(n) % 10000007);
    }
    return 0;
}

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains seven integers, a, b, c, d, e, f and n. All integers will be non-negative and 0 ≤ n ≤ 10000 and the each of the others will be fit into a 32-bit integer.
Output

For each case, print the output of the given code. The given code may have integer overflow problem in the compiler, so be careful.
Sample Input

Output for Sample Input

5

0 1 2 3 4 5 20

3 2 1 5 0 1 9

4 12 9 4 5 6 15

9 8 7 6 5 4 3

3 4 3 2 54 5 4

Case 1: 216339

Case 2: 79

Case 3: 16636

Case 4: 6

Case 5: 54

很明显可以构造出一个6*6的矩阵

⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜111111100000010000001000000100000010⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟

然后矩阵快速幂就行

/************************************************************************* > File Name: LightOJ1006.cpp > Author: ALex > Mail: [email protected] > Created Time: 2015年06月03日 星期三 21时17分35秒 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

static const int mod = 10000007;

class MARTIX {
    public:
        LL mat[6][6];
        MARTIX();
        MARTIX operator * (const MARTIX &ret)const;
        MARTIX& operator = (const MARTIX &ret);
};

MARTIX :: MARTIX() {
    memset(mat, 0, sizeof(mat));
}

MARTIX MARTIX :: operator * (const MARTIX &ret) const {
    MARTIX ans;
    for (int i = 0; i < 6; ++i) {
        for (int j = 0; j < 6; ++j) {
            for (int k = 0; k < 6; ++k) {
                ans.mat[i][j] += this-> mat[i][k] * ret.mat[k][j];
                ans.mat[i][j] %= mod;
            }
        }
    }
    return ans;
}

MARTIX& MARTIX :: operator = (const MARTIX &ret) {
    for (int i = 0; i < 6; ++i) {
        for (int j = 0; j < 6; ++j) {
            this -> mat[i][j] = ret.mat[i][j];
        }
    }
    return *this;
}

MARTIX fastpow(MARTIX A, int cnt) {
    MARTIX ans;
    for (int i = 0; i < 6; ++i) {
        ans.mat[i][i] = 1;
    }
    while (cnt) {
        if (cnt & 1) {
            ans = ans * A;
        }
        cnt >>= 1;
        A = A * A;
    }
    return ans;
}

LL f[10];

int main() {
    int t, icase = 1;
    MARTIX Base;
    for (int i = 0; i < 6; ++i) {
        Base.mat[i][0] = 1;
    }
    for (int i = 1; i < 6; ++i) {
        Base.mat[i - 1][i] = 1;
    }
    scanf("%d", &t);
    while (t--) {
        int n;
        for (int i = 0; i < 6; ++i) {
            scanf("%lld", &f[i]);
            f[i] %= mod;
        }
        scanf("%d", &n);
        if (n <= 5) {
            printf("Case %d: %lld\n", icase++, f[n]);
            continue;
        }
        MARTIX ret;
        for (int i = 0; i < 6; ++i) {
            ret.mat[0][i] = f[5 - i];
        }
        MARTIX B = fastpow(Base, n - 5);
        B = ret * B;
        printf("Case %d: %lld\n", icase++, B.mat[0][0]);
    }
    return 0;
}