HDU 1005 Number Sequence（矩阵快速幂）

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 129366    Accepted Submission(s): 31498


Problem Description 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

 

题目大意：

f(1)=f(2)=1,f(n)=(A*f(n-1)+B*f(n-2))%7,输入A，B和n，输出f(n)。

解题思路：

1 <= n <= 100,000,000，所以不能从3循环到n计算f(n)。f(3)=a+b,f(4)=a+ab+b^2，观察可以发现

所以每次只需要计算的(n-2)次方。

其实斐波那契数列也属于这种数列，只是a=1，b=1。

参考代码：

#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=1e6+50;
struct matrix
{
    int a[2][2];
    matrix()
    {
        memset(a,0,sizeof(a));
    }
};

matrix multi(matrix x,matrix y)
{
    matrix ans;
    for(int i=0; i<2; i++)
        for(int j=0; j<2; j++)
            for(int k=0; k<2; k++)
                ans.a[i][j]=(ans.a[i][j]+x.a[i][k]*y.a[k][j])%7;
    return ans;
}

int fastmod(matrix base,int n,int mod)
{
    matrix ans=base;
    ans.a[0][0]=ans.a[1][1]=1;
    ans.a[0][1]=ans.a[1][0]=0;
    while(n)
    {
        if(n&1)
            ans=multi(ans,base);
        base=multi(base,base);
        n>>=1;
    }
    return (ans.a[0][0]+ans.a[0][1])%mod;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
    int x,y,n;
    while(scanf("%d%d%d",&x,&y,&n)&&!(x==0&&y==0&&n==0))
    {
        if(n==1||n==2)
        {
            printf("1\n");
            continue;
        }
        matrix t;
        t.a[0][0]=x;
        t.a[0][1]=y;
        t.a[1][0]=1;
        t.a[1][1]=0;
        printf("%d\n",fastmod(t,n-2,7));
    }
    return 0;
}