1.原理:
不做过多解释,不会去翻高中数学教材
2.重要结论:
1.前n项和 可以 用 n + 2项的数减一得到
计算方法:
矩阵快速幂!!!
上板子题:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 21213 | Accepted: 14511 |
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
#include <vector> #include <iostream> using namespace std; typedef long long ll; typedef vector<ll> vec; typedef vector<vec> mat; const ll mod = 10000; mat mul(mat & A, mat& B) { mat C(A.size(), vec(B[0].size())); for (int i = 0; i < A.size(); ++i) for (int k = 0; k < B.size(); ++k) if (A[i][k]) for (int j = 0; j < B[0].size(); ++j) C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod; return C; } mat Pow(mat A, ll n) { mat B(A.size(), vec(A.size())); for (int i = 0; i < A.size(); ++i) B[i][i] = 1; for (; n; n >>= 1, A = mul(A, A)) if(n & 1) B = mul(B, A); return B; } int main (){ mat A(2); A.resize(2);//r行 for (int k = 0; k < 2; ++k){ A[k].resize(2);//每行为c列 } A[0][0] = 1; A[0][1] = 1; A[1][0] = 1; A[1][1] = 0; ll n; while(cin >> n && n != -1) { if (n) { mat C = Pow(A, n - 1); cout << C[0][0] << endl; } else cout << 0 << endl; } }