HDU - 1005 Number Sequence（矩阵快速幂）

Number Sequence
A number sequence is defined as follows:
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).

Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

Output
For each test case, print the value of f(n) on a single line.

Sample Input
1 1 3
1 2 10
0 0 0

Sample Output
2
5

题意：给了一个递推式f(n) = A * f(n - 1) + B * f(n - 2) ，已知这个递推式的第一项和第二项都为1，给出A，B的值，需要求这个递推式的第n项取余7。

思路：看到递推式就可以想到能用矩阵快速幂解决，利用递推式找出满足的关系矩阵：

最后记得取余一下7！

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <deque>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;

#define IOS                      \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0)
#define lowbit(a) (a & (-a))
#define ll long long
const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 10;
const int minn = 1e3 + 10;
const double PI = acos(-1.0);
#define mod 1000000007
#define eps 1e-10
/*------------------------------------------------------*/

int N = 2;

struct Matrix
{
    int mp[2][2];
    Matrix()
    {
        memset(mp, 0, sizeof(mp));
    }
    void init()
    {
        memset(mp, 0, sizeof(mp));
        for (int i = 0; i < N; i++)
            mp[i][i] = 1;
    }
};

Matrix mul(Matrix a, Matrix b)
{
    Matrix ans;
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            for (int k = 0; k < N; k++)
                ans.mp[i][j] = ((ans.mp[i][j] % 7) + ((a.mp[i][k] % 7) * (b.mp[k][j] % 7)) % 7) % 7;
    return ans;
}

Matrix mat_pow(Matrix a, int b)
{
    Matrix ans;
    ans.init();
    while (b)
    {
        if (b & 1)
            ans = mul(ans, a);
        a = mul(a, a);
        b >>= 1;
    }
    return ans;
}

int main()
{
    int a, b, n;
    while (~scanf("%d %d %d", &a, &b, &n))
    {
        if (a == 0 && b == 0 && n == 0)
            break;
        if (n == 1 || n == 2)
            printf("1\n");
        else
        {
            Matrix cnt;
            cnt.init();
            cnt.mp[0][0] = a;
            cnt.mp[0][1] = b;
            cnt.mp[1][0] = 1;
            cnt.mp[1][1] = 0;
            cnt = mat_pow(cnt, n - 2);
            printf("%d\n", (cnt.mp[0][0] + cnt.mp[0][1]) % 7);
        }
    }
    return 0;
}

本文地址：https://blog.csdn.net/weixin_44137005/article/details/110629846