hdu 3117矩陣+斐波那契數列

life4711發表於2014-07-21
矩陣

http://acm.hdu.edu.cn/showproblem.php?pid=3117

Fibonacci Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1720    Accepted Submission(s): 681


Problem Description
The Fibonacci sequence is the sequence of numbers such that every element is equal to the sum of the two previous elements, except for the first two elements f0 and f1 which are respectively zero and one.

What is the numerical value of the nth Fibonacci number?
 

Input
For each test case, a line will contain an integer i between 0 and 108 inclusively, for which you must compute the ith Fibonacci number fi. Fibonacci numbers get large pretty quickly, so whenever the answer has more than 8 digits, output only the first and last 4 digits of the answer, separating the two parts with an ellipsis (“...”). 

There is no special way to denote the end of the of the input, simply stop when the standard input terminates (after the EOF).
 

Sample Input
0
1
2
3
4
5
35
36
37
38
39
40
64
65





 



Sample Output




0
1
1
2
3
5
9227465
14930352
24157817
39088169
63245986
1023...4155
1061...7723
1716...7565





題目大意：求一個斐波那契數列，如果其位數小於等於8，那麼直接輸出該數值，如果數位大於8，輸出前四位和後四位，中間用“...”表示



大體思路：本題考查了斐波那契數列的運算，當然當n很小的情況下可以運用for迴圈求出，如果一旦n爆掉long long 那麼我們對於本題該怎麼求呢，首先，介紹一下對於斐波那契數列前幾位數的求法











而對於求後幾位我們完全可以利用矩陣來求，不能直接取模複雜度必須降到lg（n）。




#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int a[100];
const int MAX=2;
struct Matrix
{
    long long m[MAX][MAX];
};
Matrix P={0,1,
          1,1};
Matrix I={1,0,
          0,1};
Matrix matrixmul(Matrix a,Matrix b)
{
    int i,j,k;
    Matrix c;
    for(i=0;i<MAX;i++)
        for(j=0;j<MAX;j++)
        {
           c.m[i][j]=0;
           for(k=0;k<MAX;k++)
            c.m[i][j]+=(a.m[i][k]*b.m[k][j])%10000;
           c.m[i][j]%=10000;
        }
    return c;
}
Matrix quickpow(long long n)
{
    Matrix m=P,b=I;
    while(n>=1)
    {
        if(n&1)
            b=matrixmul(b,m);
        n=n>>1;
        m=matrixmul(m,m);
    }
    return b;
}
int main()
{
    int n;
    a[0]=0;
    a[1]=1;
    int temp;
    for(int i=2;;i++)
        if(a[i-1]+a[i-2]>99999999)
        {
           temp=i-1;
           break;
        }
        else
           a[i]=a[i-1]+a[i-2];
    while(~scanf("%d",&n))
    {
        Matrix tmp;
        tmp=quickpow(n);
        int ans_e=tmp.m[0][1];
        if(n<=temp)
            printf("%d\n",a[n]);
        else
        {
            int d;
            double len=0.0;
            len=log10(1.0/sqrt(5)) +(double)n*log10((1.0+sqrt(5))/2.0);
            d=(int)(pow(10,len-(int)len+3));
            printf("%d...%04d\n",d,ans_e);//將ans_e用0補齊四位
        }
    }
    return 0;
}









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