Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 13732 | Accepted: 9728 |
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:
.
这道题就是根据题意算快速幂。裸,但是这个规律比较重要了。
#include <map> #include <set> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <iostream> #include <stack> #include <cmath> #include <string> #include <vector> #include <cstdlib> //#include <bits/stdc++.h> //#define LOACL #define space " " using namespace std; typedef long long LL; typedef __int64 Int; typedef pair<int, int> paii; const int INF = 0x3f3f3f3f; const double ESP = 1e-5; const double PI = acos(-1.0); const int MOD = 1e9 + 7; const int MAXN = 100 + 10; int n, mod; LL m; struct Matrix { LL m[3][3]; int row, col; }; Matrix ori, res; void init() { memset(res.m, 0, sizeof(res.m)); ori.m[1][1] = ori.m[1][2] = ori.m[2][1] = 1; } Matrix multi(Matrix x, Matrix y) { Matrix z; memset(z.m, 0, sizeof(z.m)); for (int i = 1; i <= 2; i++) { for (int k = 1; k <= 2; k++) { for (int j = 1; j <= 2; j++) { z.m[i][j] += x.m[i][k]*y.m[k][j]%mod; } z.m[i][k] %= mod; } } return z; } Matrix pow_mod(Matrix a, LL x){ Matrix b; memset(b.m, 0, sizeof(b.m)); for (int i = 1; i <= 2; i++) { b.m[i][i] = 1; } while(x){ if(x&1) b = multi(a,b); a = multi(a, a); x >>= 1; } return b; } int main() { while (scanf("%lld", &m), m != -1) { init(); mod = 10000; res = pow_mod(ori, m); printf("%lld ", res.m[1][2]%mod); } return 0; }