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POJ 3070

編輯：關於C++

Fibonacci Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3070 Appoint description:

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_n_{? 1} + F_n_{? 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

.<喎?/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+R2l2ZW4gYW4gaW50ZWdlciA8ZW0+bjwvZW0+LCB5b3VyIGdvYWwgaXMgdG8gY29tcHV0ZSB0aGUgbGFzdCA0IGRpZ2l0cyBvZiA8ZW0+RjxzdWI+bjwvc3ViPjwvZW0+LjwvcD4KCgoKCjxwIGNsYXNzPQ=="pst"> 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 F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

Sample Output

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:

題意：題意：求第n個Fibonacci數mod(m)的結果，當n=-1時，break。其中n(where 0 ≤ n ≤ 1,000,000,000) ，m=10000；

思路：常規方法肯定超時，這道題學會了用矩陣快速冪求斐波那契。如下圖：

A = F(n - 1), B = F(N - 2)，這樣使構造矩陣的n次冪乘以初始矩陣得到的結果就是。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int inf=0x3f3f3f3f;
const int mod=10000;
struct node
{
    int mp[3][3];
}init,res;
struct node Mult(struct node x,struct node y)
{
    struct node tmp;
    int i,j,k;
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        {
            tmp.mp[i][j]=0;
            for(k=0;k<2;k++)
            {
                tmp.mp[i][j]=(tmp.mp[i][j]+x.mp[i][k]*y.mp[k][j])%mod;
            }
        }
    return tmp;
}
struct node expo(struct node x, int k)
{
    int i,j;
    struct node tmp;
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        {
            if(i==j)
                tmp.mp[i][j]=1;
            else
                tmp.mp[i][j]=0;
        }
    while(k)
    {
        if(k&1) tmp=Mult(tmp,x);
        x=Mult(x,x);
        k>>=1;
    }
    return tmp;
}
int main()
{
    int k;
    while(~scanf("%d",&k))
    {
        if(k==-1) break;
        init.mp[0][0]=1;
        init.mp[0][1]=1;
        init.mp[1][0]=1;
        init.mp[1][1]=0;
        res=expo(init,k);
        printf("%d\n",res.mp[0][1]);
    }
    return 0;
}