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uva10655

2024-09-20 04:27:19   218人阅读

Given the value of a+b and ab you will have to find the value of a n + b n Input The input file contains several lines of inputs. Each line except the last line contains 3 non-negative integers p, q and n. Here p denotes the value of a+b and q denotes the value of ab. Input is terminated by a line containing only two zeroes. This line should not be processed. Each number in the input file fits in a signed 32-bit integer. There will be no such input so that you have to find the value of 00 . Output For each line of input except the last one produce one line of output. This line contains the value of a n + b n. You can always assume that a n + b n fits in a signed 64-bit integer. Sample Input 10 16 2 7 12 3 0 0 Sample Output 68 91

矩阵快速幂。。。

很明显，要把a和b解出来很困难，因为还有虚数，讨论很烦

那么我们就要转化一下 用p和q解决问题

我们先用a^(n-1)+b^(n-1)推出a^n+b^n

要想升幂，还是生一次 只能乘a+b,但是会多出来两项（在纸上写一下，这里不方便打公式）

中间的两项提出一个a*b，变成了a^(n-2)+b^(n-2) 那么我们就可以递推了

但是太慢了，就用矩阵快速幂。。。

构造矩阵就行了 

注意特判n==0

技术分享 
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
struct mat {
    ll a[3][3];
} A, B;
ll p, q, n;
mat operator * (mat A, mat B)
{
    mat ret; memset(ret.a, 0, sizeof(ret.a));
    for(int i = 1; i < 3; ++i)
        for(int j = 1; j < 3; ++j)
            for(int k = 1; k < 3; ++k) ret.a[i][j] = ret.a[i][j] + A.a[i][k] * B.a[k][j];
    return ret;         
}
mat power(mat A, ll t)
{
    mat ret; memset(ret.a, 0, sizeof(ret.a));
    for(int i = 1; i < 3; ++i) ret.a[i][i] = 1;
    for(; t; t >>= 1, A = A * A) if(t & 1) ret = ret * A;
    return ret;
}
int main()
{
    while(scanf("%lld%lld%lld", &p, &q, &n) == 3)
    {
        if(n == 0) { puts("2"); continue; }
        if(n == 1) { printf("%lld\n", p); continue; }
        if(n == 2) { printf("%lld\n", p * p - 2 * q); continue; }
        A.a[1][1] = p; A.a[1][2] = -q;
        A.a[2][1] = 1; A.a[2][2] = 0;
        B.a[1][1] = p * p - 2 * q; B.a[2][1] = p;
        B = power(A, n - 2) * B;
        printf("%lld\n", B.a[1][1]);        
    }
    return 0;
}
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uva10655


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