HDU 1005 Number Sequence【斐波那契数列/循环节找规律/矩阵快速幂/求(A * f(n

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 187893    Accepted Submission(s): 46820

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

 【分析】：

【代码】：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int MOD = 7;
typedef vector<LL> vec;
typedef vector<vec> mat;
LL n;

mat mul(mat &A,mat &B)
{
    mat C(A.size(),vec(B[0].size()));
    for( int i = 0; i < A.size(); i++ ){
        for( int j = 0; j < B[0].size();j++ ){
            for( int k = 0; k < B.size();k++ ){
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]);
                C[i][j] %= MOD;
            }
        }
    }
    return C;
}
mat pow(mat A,LL n)
{
    mat B(A.size(),vec(A.size()));
    for( int i = 0; i < A.size(); i++ ){
        B[i][i] = 1;
    }
    while(n > 0){
        if(n & 1)B = mul(B,A);
        A = mul(A,A);
        n >>= 1;
    }
    return B;
}
void solve(LL a,LL b)
{
    mat A(2,vec(2));
    A[0][0] = a;
    A[0][1] = b;
    A[1][0] = 1;
    A[1][1] = 0;
    A = pow(A,n-2);
    printf("%lld
",(A[0][0]+A[0][1])%7);
}
int main()
{
    LL a,b;
    while(~scanf("%lld%lld%lld",&a,&b,&n),a)
        solve(a,b);
    return 0;
}