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)]

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#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; }