题目连接
http://poj.org/problem?id=3070
Fibonacci
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
裸的矩阵快速幂。。
#include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> #include<vector> #include<map> using std::min; using std::sort; using std::pair; using std::swap; using std::vector; using std::multimap; #define pb(e) push_back(e) #define sz(c) (int)(c).size() #define mp(a, b) make_pair(a, b) #define all(c) (c).begin(), (c).end() #define iter(c) __typeof((c).begin()) #define cls(arr, val) memset(arr, val, sizeof(arr)) #define cpresent(c, e) (find(all(c), (e)) != (c).end()) #define rep(i, n) for(int i = 0; i < (int)n; i++) #define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i) const int M = 10000; const int N = 1000000; const int INF = 0x3f3f3f3f; struct Matrix { typedef vector<int> vec; typedef vector<vec> mat; inline mat mul(mat &A, mat &B) { mat C(sz(A), vec(sz(B[0]))); rep(i, sz(A)) { rep(k, sz(B)) { rep(j, sz(B[0])) { C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M; } } } return C; } inline mat pow(mat &A, int n) { mat ret(sz(A), vec(sz(A[0]))); rep(i, sz(A)) ret[i][i] = 1; while(n) { if(n & 1) ret = mul(ret, A); A = mul(A, A); n >>= 1; } return ret; } inline void solve(int n) { mat ans(2, vec(2)); ans[0][0] = 1, ans[0][1] = 1; ans[1][0] = 1, ans[1][1] = 0; ans = pow(ans, n); printf("%d ", ans[1][0]); } }go; int main() { #ifdef LOCAL freopen("in.txt", "r", stdin); freopen("out.txt", "w+", stdout); #endif int n; while(~scanf("%d", &n) && ~n) { go.solve(n); } return 0; }