题目链接:点我
题意:
给你a + b 和 a * b 让你计算 $ a ^{n}$ + $ b ^{n}$
思路:
矩阵快速幂.
刚开始看见这个式子我也是懵逼的,后来就想,看看能不能找规律,于是自己手写了几项, 发现还是真的有一部份规律. $ a^{n}$ + bn =($ a^{n - 1}$ + bn−1) * ( a + b - a * b ) ,突然发现自己乱写乱画也弄出个公式(高兴), 于是愉快的A 了这题.
递推矩阵 😒 \begin {pmatrix} q & -p\ 1 & 0\ \end{pmatrix}$
代码:
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| #include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long LL;
struct mat{
LL a[10][10];
mat(){memset(a, 0, sizeof(a)); }
mat operator *( const mat q){
mat c;
for(int i = 1; i <= 2;++i)
for(int k = 1; k<= 2; ++k)
for(int j =1; j<= 2; ++j)
c.a[i][j] += a[i][k] * q.a[k][j];
return c;
}
};
mat qpow(mat x,LL n){
mat ans;
ans.a[1][1] = ans.a[2][2] = 1;
while(n){
if(n&1) ans =ans * x;
x =x * x;
n >>= 1;
}return ans;
}
int main(){
LL p, q, n;
while(scanf("%lld %lld %lld", &q, &p, &n) == 3){
if(n == 0){
printf("2\n");
continue;
}else if(n == 1){
printf("%lld\n",q);
continue;
} else if(n == 2){
printf("%lld\n",q*q - 2*p);
continue;
}mat ans;
ans.a[1][1] = q;
ans.a[1][2] = -p;
ans.a[2][1] = 1;
ans = qpow(ans,n-2);
LL sum1 = q;
LL sum2 = q*q - 2 * p;
LL sum = sum1 * ans.a[1][2] + sum2 * ans.a[1][1];
printf("%lld\n",sum);
}return 0;
}
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