POJ 3070 Fibonacci——————矩陣快速冪解法
**Fibonacci ** Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 19481 Accepted: 13467 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:
Source
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <cmath> #include <map> #include <set> #include <stack> #include <queue> using namespace std; typedef long long ll; typedef unsigned long long ull; const int INF = 0x3f3f3f3f; const int MAXN = 17; const double PI = acos(-1.0); const int mod = 10000; int N; ll b_n = 0; ll C[MAXN]; ll h[MAXN]; struct matrix{ ll m[MAXN][MAXN];}; matrix multi(matrix a,matrix b)//矩陣乘法 { matrix tmp; for(int i=1;i<=N;i++) for(int j=1;j<=N;j++) { tmp.m[i][j] = 0; for(int k=1;k<=N;k++) tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k]*b.m[k][j] )%mod; } return tmp; } matrix pow_mod(matrix a,int n)//矩陣快速冪 { matrix res; for(int i=1;i<=N;i++) for(int j=1;j<=N;j++) res.m[i][j] = (i==j); while(n) { if(n&1) res = multi(res,a); a = multi(a,a); n>>=1; } return res; } void init(matrix &res,matrix &H)//初始化 { for(int i=1;i<=N;i++) res.m[1][i]=C[i];//第一列賦值 res.m[1][N] = b_n; for(int i=2;i<=N;i++)//構造單位矩陣 for(int j=1;j<=N;j++) res.m[i][j] = (i==j+1); res.m[N][N-1] = 0; res.m[N][N] = 1; for(int i=1;i<=N;i++) { H.m[i][1] = h[N-i]; for(int j=2;j<=N;j++) H.m[i][j] = 0; } H.m[N][1]=1; } void slove(int k,int n) { matrix res,H; init(res,H); res = pow_mod(res,n-k+1); res = multi(res,H); ll ans = res.m[2][1]; printf("%lld\n",ans); } int main() { int k,n; k=2; C[1]=1; C[2]=1; h[1]=1; h[2]=1; while(~scanf("%d",&n)&& -1!=n) { if(!n) { printf("0\n"); continue; } N = k+1; slove(k, n); } return 0; }