Problem E
Contemplation! Algebra
Input: Standard Input

Output: Standard Output

Time Limit: 1 Second

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 andq 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+bⁿ fits in a signed 64-bit integer.

            Sample Input                                                                               Output for Sample Input

10 16 2 

7 12 3 

0 0

题意：已知p=a+b;q=a*b;求a^n+b^n=ans? 注意a,b是实数哦！还有题目说不用考虑0^0这种情况！

那么分析一下：n = 0 ,　　ans₀ = 2 ;

　　　　　　　n = 1 ,　　ans₁ = a + b = p ;

　　　　　　　n = 2 ,　　ans₂ = p * p - 2 * q = ans₁ * p - ans₀ * q;

　　同理　　　n = 3 ,　　ans₃ = ans₂ * p - ans₁ * q;　　　　

　　　　　　　n = 4 ,       .......

　　　　　　　..................

　　　　　　　所以矩阵为

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1596

转载请注明出处：寻找&星空の孩子

对了这个题还有一点比较坑爹；就是最后那组测试数据。。。不解释我错了5次。。。

#include<cstring>//用c++的输入就过了。
#include<cstdio>
#include<iostream>
using namespace std;

#define LL long long

struct matrix
{
    LL mat[2][2];
};

matrix multiply(matrix a,matrix b)
{
    matrix c;
    memset(c.mat,0,sizeof(c.mat));
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            if(a.mat[i][j]==0)continue;
            for(int k=0;k<2;k++)
            {
                if(b.mat[j][k]==0)continue;
                c.mat[i][k]+=a.mat[i][j]*b.mat[j][k];
            }
        }
    }
    return c;
}

matrix quickmod(matrix a,LL m)
{
    matrix res;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            res.mat[i][j]=(i==j);
    while(m)
    {
        if(m&1) res=multiply(res,a);
        m>>=1;
        a=multiply(a,a);
    }
    return res;
}

int main()
{
    LL p,q,n;
 //   while(scanf("%lld%lld%lld",&p,&q,&n),p+q+n)
    while(cin>>p>>q>>n)
    {
  //      if(!n&&(!p||!q)) break;

 //       scanf("%lld",&n);
        if(n==0)printf("2\n");//cout<<2<<endl;//
        else if(n==1)printf("%lld\n",p);// cout<<p<<endl;//
        else if(n==2) printf("%lld\n",p*p-2*q);//cout<<p*p-2*q<<endl;//
        else
        {
            //初始矩阵（p*p-2*q,p）
            matrix ans;
            ans.mat[0][0]=p;
            ans.mat[0][1]=1;
            ans.mat[1][0]=-q;
            ans.mat[1][1]=0;

            ans=quickmod(ans,n-1);
            LL ant=p*ans.mat[0][0]+2*ans.mat[1][0];
 //           cout<<ant<<endl;
            printf("%lld\n",ant);
//            printf("%lld\n",(p*ans.mat[0][0]+2*ans.mat[1][0]));
        }
    }
    return 0;
}

加油，少年！！！