【UVA10655】 Contemplation! Algebra

题目

给定 \(p = a + b\) 和 \(q = ab\) 和 \(n\)，求 \(a ^ n + b ^ n\)。

$0\le n\lt 2^{63} $

分析

大水题。

先考虑 \(n\) 较小的情况，可以很容易的想到递推：
\[
\begin{array}{}
\text{令} F(i) & = a ^ n + b ^ n \ & = (a + b)(a ^ {n - 1} + b ^ {n - 1}) - (ab ^ {n - 1} + a^{n - 1}b) \ & = (a + b)(a ^ {n - 1} + b ^ {n - 1}) - ab(a ^ {n - 2} + b ^ {n - 2}) \ & = p \times F(i - 1) - q \times F(i - 2)
\end{array}
\]
然后发现这个递推式可以用矩阵优化：
\[
\left[\begin{matrix}
p & - q \ 1 & 0
\end{matrix}\right]
\times
\left[\begin{matrix}
F[i] \ F[i - 1]
\end{matrix}\right] =
\left[\begin{matrix}
F[i] \times p & + & F[i - 1]\times (-q) \ F[i] \times 1 & + & F[i - 1]\times 0
\end{matrix}\right] =
\left[\begin{matrix}
F[i + 1] \ F[i]
\end{matrix}\right]
\]
即：
\[
\left[\begin{matrix}
p & - q \ 1 & 0
\end{matrix}\right]^n
\times
\left[\begin{matrix}
F[1] \ F[0]
\end{matrix}\right] =
\left[\begin{matrix}
F[n + 1] \ F[n]
\end{matrix}\right]
\]
显然，\(F[1] = p,\ F[0] = 2\)。

代码

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAXN = 10;

struct matrix {
    ll a[MAXN][MAXN]; int rowSize, lineSize;

    matrix(int x, int y) {
        rowSize = x; lineSize = y;
    }

    ll *operator [](const unsigned &i) {return a[i];}

    matrix operator *(matrix y) {
        matrix ans(rowSize, y.lineSize);
        for(int i = 0; i < ans.rowSize; i++)
            for(int j = 0; j < ans.lineSize; j++) {
                ans[i][j] = 0;
                for(int k = 0; k < lineSize; k++)
                    ans[i][j] += a[i][k] * y[k][j];
            }
        return ans;
    }
} u(2, 2);

matrix qPow(matrix x, ll b) {
    matrix ans = u, base = x;
    while(b) {
        if(b & 1)
            ans = ans * base;
        base = base * base;
        b >>= 1;
    }
    return ans;
}

int main() {
    ios::sync_with_stdio(false);
    u[0][0] = 1; u[0][1] = 0;
    u[1][0] = 0; u[1][1] = 1;
    ll p, q, n;
    while(scanf("%lld%lld%lld", &p, &q, &n) == 3) {
        matrix a(2, 2), b(2, 2), st(2, 1);
        a[0][0] = p; a[0][1] = -q;
        a[1][0] = 1; a[1][1] = 0;
        st[0][0] = p; st[1][0] = 2;
        b = qPow(a, n) * st;
        printf("%lld\n", b[1][0]);
    }
    return 0;
}

UVA10655 Contemplation! Algebra

原文地址：https://www.cnblogs.com/zhylj/p/9931671.html