链接：http://poj.org/problem?id=3070
- Fibonacci
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 16351 Accepted: 11470
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:
.
.
#include <iostream>
#include <string.h>using namespace std;
const int maxn =15;
const int mod =10000;
struct matrix
{long long int a[2][2];
};
matrix multply(matrix x,matrix y)
{int i,j,k;matrix result;memset(result.a,0,sizeof(result.a));//**for (k=0; k<2; k++)for (i=0; i<2; i++){if(x.a[i][k])for (j=0; j<2; j++){if (y.a[k][j]){result.a[i][j]+=x.a[i][k]*y.a[k][j];if (result.a[i][j]>=mod)result.a[i][j]%=mod;}}}return result;
}
void matrix_pow(matrix a,int nn)
{matrix result;memset(result.a,0,sizeof(result.a));int i,j,k;for (i=0; i<2; i++)result.a[i][i]=1;while (nn){if (nn&1)result=multply(result,a);nn>>=1;a=multply(a,a);
// int i,j;
// for (i=0; i<2; i++)
// for (j=0; j<2; j++)
// if(!j)
// cout<<a.a[i][j]<<" ";
// else
// cout <<a.a[i][j]<<endl;}cout<<result.a[0][1]<<endl;}int main ()
{long long int nn;while (cin>>nn){if (nn==-1)break;matrix a;a.a[0][0]=a.a[0][1]=a.a[1][0]=1;a.a[1][1]=0;
// int i,j;
// for (i=0; i<2; i++)
// for (j=0; j<2; j++)
// if(!j)
// cout<<a.a[i][j]<<" ";
// else
// cout <<a.a[i][j]<<endl;matrix_pow(a,nn);}return 0;
}