题目链接：UVA 10655 Contemplation! Algebra
题目大意：
已知(a+b)和(ab)，求(a^n+b^n)。
题解：
递推式不难推出来。
构造矩阵：
[left(egin{matrix}
a^i+b^i \
a^{i-1}+b^{i-1}
end{matrix}
ight)
=
left(egin{matrix}
a+b & -ab \
1 & 0
end{matrix}
ight)
imes
left(egin{matrix}
a^{i-1}+b^{i-1} \
a^{i-2}+b^{i-2}
end{matrix}
ight)
]
所以：
[left(egin{matrix}
a^n+b^n \
a^{n-1}+b^{n-1}
end{matrix}
ight)
=
left(egin{matrix}
a+b & -ab \
1 & 0
end{matrix}
ight)^{n-1}
imes
left(egin{matrix}
a+b \
2
end{matrix}
ight)
]
另外，题目规定一行只有两个(0)时结束，但是存在一行有三个数但前两个是(0)的情况，所以不能单纯判断前两个数是不是(0)。
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define ll long long
struct Matrix { // 矩阵
int row, col;
ll 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]);
return temp;
}
Matrix MatrixFastPow(Matrix base, ll 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() {
ll p, q, n;
Matrix base;
base.row = base.col = 2;
base.num[1][0] = 1;
base.num[1][1] = 0;
while (cin >> p >> q >> n) {
if (!n) {
cout << 2 << endl;
continue;
}
base.num[0][0] = p;
base.num[0][1] = -q;
Matrix ans = MatrixFastPow(base, n - 1);
cout << ans.num[0][0] * p + ans.num[0][1] * 2 << endl;
}
return 0;
}