HDU 1005 Number Sequence

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

看了这个数据，n的值有点大，第一反应就是矩阵快速幂，不过AC之后才感觉不用那么复杂，不过还是矩阵快速幂做比较放心

#include <iostream>
#include <cstdio>
using namespace std;
struct Mat
{
    int matrix[2][2];
};
Mat Multi(const Mat& a, const Mat& b)
{
int i, j, k;
Mat c;
for (i = 0; i < 2; i++)
{
   for (j = 0; j < 2; j++)
   {
    c.matrix[i][j] = 0;
    for (k = 0;k < 2; k++)
     c.matrix[i][j] += a.matrix[i][k] * b.matrix[k][j] % 7;
    c.matrix[i][j] %= 7;
   }
}
return c;
}int main()
{
    int A, B, n;
    int f[3];
    while(scanf("%d%d%d",&A,&B,&n)!=EOF && !(A==0 && B==0 && n==0))
    {
        Mat stand = {0, 1, B, A};
        Mat e = {1, 0, 0, 1};
        f[1]=1;f[2]=1;
         if (n <= 2)
         {
              printf("%d\n", f[n]);
              continue;
         }
         Mat ans = e;
         Mat tmp = stand;
         n = n - 2;
         while(n)
         {
            if (n & 1)
            ans = Multi(ans, tmp);
            tmp = Multi(tmp, tmp);
            n >>= 1;
         }
         printf("%d\n", (ans.matrix[1][0] * f[1] + ans.matrix[1][1] * f[2]) % 7);
     }
return 0;
}