HDU - 1005 -Number Sequence(矩阵快速幂系数变式)

A number sequence is defined as follows: 

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. 

Given A, B, and n, you are to calculate the value of f(n).

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

Output

For each test case, print the value of f(n) on a single line.

Sample Input

1 1 3
1 2 10
0 0 0

Sample Output

2
5

题解：把系数矩阵的数换下即可

代码：

#include<stdio.h>
#include<string.h>
#define MAX 10
typedef long long ll;

ll p,q,MOD=7;
struct mat{
    ll a[MAX][MAX];
};

mat operator *(mat x,mat y)   //重载*运算
{
    mat ans;
    memset(ans.a,0,sizeof(ans.a));
    for(int i=1;i<=2;i++){
        for(int j=1;j<=2;j++){
            for(int k=1;k<=2;k++){
                ans.a[i][j]+=x.a[i][k]*y.a[k][j];
                ans.a[i][j]%=MOD;
            }
        }
    }
    return ans;
}
mat qsortMod(mat a,ll n)   //矩阵快速幂
{
    mat t;
    t.a[1][1]=p;t.a[1][2]=q;  //变式的系数
    t.a[2][1]=1;t.a[2][2]=0;
    while(n){
        if(n&1) a=t*a;   //矩阵乘法不满足交换律，t在前
        n>>=1;
        t=t*t;
    }
    return a;
}
int main()
{
    ll n;
   while(scanf("%lld%lld%lld",&p,&q,&n)!=EOF)
   {
    if(p==0&&q==0&&n==0)
    break;
    if(n==1) printf("1\n");
    else if(n==2) printf("1\n");
    else{
        mat a;
        a.a[1][1]=1;a.a[1][2]=0;
        a.a[2][1]=1;a.a[2][2]=0;
        a=qsortMod(a,n-2);
        printf("%lld\n",a.a[1][1]);
    }
   }
    return 0;
}