poj3070 Fibonacci 矩陣快速冪
Fibonacci
Time Limit: 1000MS | Memory Limit: 65536K |
Total Submissions: 19733 | Accepted: 13621 |
Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is
Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
0 9 999999999 1000000000 -1
Sample Output
0 34 626 6875
Hint
As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by
Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:
思路:
矩陣快速冪裸題
程式碼:
#include<cstdio>
#include<cstring>
using namespace std;
#define ll long long
#define N 2
const ll mod = 10000;
struct Matrix
{
int n;
ll d[N][N];
void init(int n)
{
this -> n = n;
memset(d,0,sizeof(d));
}
Matrix operator *(Matrix b)
{
Matrix ans;
ans.init(n);
for (int i = 0;i < n;i ++)
for (int j = 0;j < n;j ++)
for (int k = 0;k < n;k ++)
ans.d[i][j]=(ans.d[i][j]+d[i][k]*b.d[k][j])%mod;
return ans;
}
};
ll a[N][N] = {{0,1},
{1,1}};
Matrix quick(Matrix a,ll b)
{
Matrix ans;
ans.init(a.n);
for (int i = 0;i < ans.n;i ++)
ans.d[i][i] = 1;
while (b)
{
if (b & 1) ans = ans * a;
a = a * a;
b >>= 1;
}
return ans;
}
int main()
{
ll n;
while (~scanf("%lld",&n) && n != -1)
{
if (n < 2)
{
printf("%lld\n",n);
continue;
}
Matrix ans,x;
x.init(2),ans.init(2);
for (int i = 0;i < N;i ++)
for (int j = 0;j < N;j ++)
x.d[i][j] = a[i][j];
x = quick(x,n - 1);
ans.d[0][0] = 0,ans.d[0][1] = 1;
ans = ans * x;
printf("%lld\n",ans.d[0][1]);
}
return 0;
}