[HDOJ1005]Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 124651    Accepted Submission(s): 30286

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

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

 

　　这题可以找规律，但是首先想到的还是矩阵快速幂。

 

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;

#define MOD 7
#define MAXN 10

typedef struct MAT
{
    int d[MAXN][MAXN];
    int r, c;
    MAT() 
    {
        r = c = 0;
        memset(d, 0, sizeof(d));
    }
}MAT;

MAT mul(MAT m1, MAT m2, int mod)
{
    MAT ans = MAT();
    ans.r = m1.r;
    ans.c = m2.c;
    for(int i = 0; i < m1.r; i++)
    {
        for(int j = 0; j < m2.r; j++)
        {
            if(m1.d[i][j])
            {
                for(int k = 0; k < m2.c; k++)
                {
                    ans.d[i][k] = (ans.d[i][k] + m1.d[i][j] * m2.d[j][k]) % mod;
                }
            }
        }
    }
    return ans;
}

MAT quickmul(MAT m, int n, int mod)
{
    MAT ans = MAT();
    for(int i = 0; i < m.r; i++)
    {
        ans.d[i][i] = 1;
    }
    ans.r = m.r;
    ans.c = m.c;
    while(n)
    {
        if(n & 1)
        {
            ans = mul(m, ans, mod);
        }
        m = mul(m, m, mod);
        n >>= 1;
    }
    return ans;
}

int main() {
    int n, m, t, a, b, k;
    while(scanf("%d %d %d", &a, &b, &n) && a + b + n) {
        MAT tmp, ans;
        tmp.r = 2, tmp.c = 2;
        ans.r = 2, ans.c = 1;
        if(n == 1 || n == 2) {
            printf("%d
", 1);
        }
        else {
            tmp.d[0][0] = a;
            tmp.d[0][1] = b;
            tmp.d[1][0] = 1;
            tmp.d[1][1] = 0;
            ans = quickmul(tmp, n - 2, MOD);
            printf("%d
", (ans.d[0][0]+ans.d[0][1]) % MOD);
        }
    }
}