Fibonacci
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9156 | Accepted: 6494 |
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:
.
这个题是要用矩阵的知识解决斐波那契数列，题目中给的公式很简单，直接往上套就行了，那我们就先看除了题目给的方法外的另一个方法，当然还是矩阵。
先看下面的公式：
不难看出这个式子是把：
写成矩阵的形式而已。
将上面的式子迭代可得：
故可得一般式为：
继续展开会得到：
设：
则：
所以：
现在已经很明显了，其中还要用到矩阵快速幂的知识，代码如下：
#include <stdio.h>
#include <string.h>
#include <math.h>
typedef __int64 int64;
int64 c[2][2],ans[2][2],d[2][2],a[2][2];
int main()
{
int64 i,j,n,k,t;
while(scanf("%I64d",&n)!=EOF)
{
if(n==-1)
break;
if(n==0)
{
printf("0\n");
continue;
}
if(n==1)
{
printf("1\n");
continue;
}
memset(ans,0,sizeof(ans));
for(i=0;i<2;i++)
ans[i][i]=1;
a[0][0]=a[0][1]=a[1][0]=1;
a[1][1]=0;
n=n-2;
while(n!=0)
{
if(n%2==1)
{
memset(d,0,sizeof(d));
for(i=0;i<2;i++)
for(j=0;j<2;j++)
if(a[i][j])
for(k=0;k<2;k++)
{
d[i][k]+=ans[i][j]*a[j][k];
d[i][k]=d[i][k]%10000;
}
for(i=0;i<2;i++)
for(j=0;j<2;j++)
ans[i][j]=d[i][j];
}
memset(c,0,sizeof(c));
for(i=0;i<2;i++)
for(j=0;j<2;j++)
{
if(a[i][j]==0)
continue;
for(k=0;k<2;k++)
{
c[i][k]+=a[i][j]*a[j][k];
c[i][k]=c[i][k]%10000;
}
}
for(i=0;i<2;i++)
for(j=0;j<2;j++)
a[i][j]=c[i][j];
n=n/2;
}
t=(ans[0][0]+ans[0][1])%10000;
if(t==0)
printf("0\n");
else
printf("%I64d\n",t);
}
return 0;
}
快速Fibonacci数列，矩阵法