UVA 10655 Contemplation! Algebra

Given the value of a+b and ab you will have to find the value of aⁿ+bⁿ

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 aⁿ+bⁿ.  You can always assume that aⁿ+bⁿfits in a signed 64-bit integer.

 

可以推出a ^ n + b ^ n = (a + b)[a ^ ( n - 1) + b ^ (n - 1)] - ab[a ^ (n - 2) + b ^ (n - 2)]

    #include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define LL long long
struct matrix
{
    LL f[2][2];
};
matrix mul(matrix a,matrix b)
{
    LL i,j,k;
    matrix c;
    memset(c.f,0,sizeof(c.f));
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
            for(k=0;k<2;k++)
            c.f[i][j]+=a.f[i][k]*b.f[k][j];
    return c;
}
matrix ksm(matrix e,LL n)
{
    matrix s;
    s.f[0][0]=s.f[1][1]=1;
    s.f[0][1]=s.f[1][0]=0;
    while(n)
    {
        if(n&1)
            s=mul(s,e);
        e=mul(e,e);
        n=n>>1;
    }
    return s;
}
matrix e;
int main()
{
    LL p,q,n;
    while(cin>>p>>q>>n)
    {
        if(n==0)
        {
            cout<<2<<endl;
            continue;
        }
        e.f[0][0]=p;e.f[0][1]=1;
        e.f[1][0]=-q;e.f[1][1]=0;
        e=ksm(e,n-1);
        LL ans;
        ans=p*e.f[0][0]+2*e.f[1][0];
        cout<<ans<<endl;
    }
    return 0;
}