hust 1333

题目描述

We often encounter sequence like this, 0, 1, 1, 2, 3, 5, 8, ... , which is called Fibonacci sequence. It's defined by the recurrence F[0] = 0 F[1] = 1 F[n] = F[n-1] + F[n-2], for n > 1. Now you need to calculate the value of the kth term mod 1000000009 of the sequence defined by the recurrence below. G[0] = u G[1] = v G[n] = a * G[n-1] + b * G[n-2], for n > 1.

输入

The input has 30000 cases, one line per case. Each line contains five integers, u, v, a, b, k. 0<=u, v, a, b,k<=1000000009.

输出

The value of G[k] mod 1000000009, one line per case.

样例输入

0 1 1 1 7
3 5 2 7 6
1 2 3 4 5

样例输出

13
5879
614
又是一道矩阵快速幂的问题，在我们hust上，快速幂的题目还有不少啊！

#include <iostream>
#include <cstdio>
using namespace std;
const long long mod=1000000009;
struct Mat
{
    long long  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] % mod;
    c.matrix[i][j] %= mod;
   }
}
return c;
}int main()
{
    long long  a, b, f[2];
    int n;
    //Mat stand = {0, 1, 0, 0, 0, 1, a3, a2, a1};
    Mat e = {1, 0, 0, 1};
    //scanf("%d", &tot);
    while(scanf("%lld%lld%lld%lld%d",&f[0],&f[1],&a,&b,&n)!=EOF)
    {
         //scanf("%d%d%d%d", &f[0], &f[1], &f[2], &n);
         Mat stand = {0, 1, b, a};
         if (n <= 1)
         {
              printf("%lld
",f[n]);
              continue;
         }
         Mat ans = e;
         Mat tmp = stand;
         n = n - 1;
         while(n)
         {
            if (n & 1)
            ans = Multi(ans, tmp);
            tmp = Multi(tmp, tmp);
            n >>= 1;
         }
         printf("%lld
", (ans.matrix[1][0] * f[0] + ans.matrix[1][1] * f[1] ) % mod);
     }
return 0;
}