Contemplation! Algebra

　　　　　　Contemplation! Algebra

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

 

附：debug：

input：

0 0 0
0 0 1
0 0 200000000
1 3 0
999999 999999 0
1 0 200000000
2 0 61
2 1 0
2 1 100
2 1 200000000
0 0 0
0 1 1
0 1 2
0 1 3
0 1 199999999
0 1 200000000
0 1 200000001
0 1 200000002
5 6 7
10 20 10
90 87 3
999 231 2
0 0

 

output：

2
0
0
2
2
1
2305843009213693952
2
2
2
2
0
-2
0
0
2
0
-2
2315
393600000
705510
997539


题意：输入p,q,n;p=a+b，q=a*b，输出a^n+b^n
TIP:令f（n）=a^n+b^n;则f(n）*(a+b)=a^(n+1)+b^(n+1)+a*b^n+b*a^n
　　　　　　　　　　　　　　　　　　　 =f(n+1)+a*b*f(n-1)
　　即：f（n+1)=-(a+b)*f(n)+a*b*f(n-1)
好了，矩阵上场，对于形如f（n+1)=p*f(n)+q*f(n-1)的；都可以转换为（p,q）
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　(1,0)…………%…………这是个2*2的矩阵

然后{f（n）,f(n-1)}……（此为1*2的矩阵）=(那个矩阵）^(n-2)*｛f(2),f(1)｝;

 

注意：如果输入0,0,2；你的程序运行不？

　　scanf（……………………==3），我的如果不写的话就超时。

 

代码:

 

///a^(n+1)+b^(n+1)=(a^n+b^n)p−(a^(n−1)+b^(n−1))
#include<string.h>
#include<stdio.h>

long long p,q,n;

struct mat
{
    long long v[2][2];
    mat(){memset(v,0,sizeof(v));}
    mat operator * (mat c)
    {
        mat ans;
        for(int i=0;i<2;i++)///矩阵乘法再加上原有的原数
        {
            for(int j=0;j<2;j++)
            {
                for(int k=0;k<2;k++)
                {
                    ans.v[i][j]=ans.v[i][j]+v[i][k]*c.v[k][j];
                }
               }
          }
          return ans;
     }
};

mat pow_mod(mat x,long long k)
{
    mat ans;
    ans.v[0][0]=ans.v[1][1]=1;

    while(k)///二分求定以后的*
    {
        if(k&1)
            ans=ans*x;
        x=x*x;
        k>>=1;
     }

     return ans;
}

long long solve()
{
    if(n==0)
        return 2;
    if(n==1)
        return p;
    if(n==2)
        return p*p-2*q;

    mat a;
    a.v[0][0]=p;
    a.v[0][1]=-q;
    a.v[1][0]=1;
    a=pow_mod(a,n-2);

    return a.v[0][1]*p+a.v[0][0]*(p*p-2*q);
}

int main()
{
    while(scanf("%lld%lld%lld",&p,&q,&n)==3)
    {
        printf("%lld\n",solve());
    }
    return 0;
}