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CF450B Jzzhu and Sequences

作者:互联网

题目链接:CF450B Jzzhu and Sequences

题目大意:

已知\(x\)和\(y\),给你\(n\),求\(f_n\%1e9+7\)。

题解:
因为\(f_i=f_{i-1} + f_{i+1}\),所以\(f_{i+1} = f_i-f_{i-1}\),即\(f_i=f_{i-1}-f_{i-2}\)。
寻找规律发现这个数列每\(6\)个数为一个循环,则直接输出\(n\%6\)之后对应的值就行了。

#include <iostream>
using namespace std;
const int mod = 1e9 + 7;

int main() {
    long long f[7], n;
    cin >> f[1] >> f[2] >> n;
    for (int i = 3; i <= 6; ++i) f[i] = f[i - 1] - f[i - 2];
    while (f[(n - 1) % 6 + 1] < 0) f[(n - 1) % 6 + 1] += mod;
    cout << (f[(n - 1) % 6 + 1]) % mod;
    return 0;
}

当然这不是正解!!!!!
正解是构造矩阵然后用矩阵快速幂求解。

\[\left(\begin{matrix} f_i \\ f_{i-1} \end{matrix}\right) = \left(\begin{matrix} 1 & -1 \\ 1 & 0 \end{matrix}\right) \times \left(\begin{matrix} f_{i-1} \\ f_{i-2} \end{matrix}\right) \]

所以

\[\left(\begin{matrix} f_n \\ f_{n-1} \end{matrix}\right) = \left(\begin{matrix} 1 & -1 \\ 1 & 0 \end{matrix}\right)^{n-2} \times \left(\begin{matrix} f_2 \\ f_1 \end{matrix}\right) \]

几个注意点:
1. 数据比较大,要开long long;
2. 快速幂里用\(n-1\)是因为\(n\)可能为\(1\),如果是\(n-2\)的话会变成负数,使快速幂无法跳出循环;
3. 结果可能为负数,取余时要加\(1e9+7\)直至大于\(0\)(加一次可能不够)。

#include <iostream>
#include <cstring>
using namespace std;
const int mod = 1e9 + 7;

struct Matrix { // 矩阵
    int row, col;
    long long num[2][2];
};

Matrix multiply(Matrix a, Matrix b) { // 矩阵乘法
    Matrix temp;
    temp.row = a.row, temp.col = b.col;
    memset(temp.num, 0, sizeof(temp.num));
    for (int i = 0; i < a.row; ++i)
        for (int j = 0; j < b.col; ++j)
            for (int k = 0; k < a.col; ++k)
                temp.num[i][j] = (temp.num[i][j] + a.num[i][k] * b.num[k][j] % mod) % mod;
    return temp;
}

Matrix fastPow(Matrix base, long long k) { // 矩阵快速幂
    Matrix ans;
    ans.row = ans.col = 2;
    ans.num[0][0] = ans.num[1][1] = 1;
    ans.num[0][1] = ans.num[1][0] = 0;
    while (k) {
        if (k & 1) ans = multiply(ans, base);
        base = multiply(base, base);
        k >>= 1;
    }
    return ans;
}

int main() {
    long long x, y, n;
    cin >> x >> y >> n;
    Matrix base;
    base.row = base.col = 2;
    base.num[0][0] = base.num[1][0] = 1;
    base.num[0][1] = -1;
    base.num[1][1] = 0;
    Matrix ans = fastPow(base, n - 1);
    cout << ((ans.num[1][0] * y + ans.num[1][1] * x) % mod + mod) % mod << endl;
    return 0;
}

标签:matrix,ans,Jzzhu,num,long,base,Sequences,CF450B,Matrix
来源: https://www.cnblogs.com/IzumiSagiri/p/14318814.html