Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 30932 | Accepted: 20284 |
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:
知识点矩阵快速幂加速递推
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const ll m = 10000;
// 矩阵的快速幂
struct matrix{ ll m[3][3]; };
matrix operator *(const matrix& a, const matrix& b) { // 重载*为矩阵乘法,注意const
matrix c;
memset(c.m,0, sizeof(c.m));
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 2; k++) {
c.m[i][j] = (c.m[i][j] + a.m[i][k] * b.m[k][j]) % m;
}
}
}
return c;
}
matrix pow_matrix(matrix a, ll n) {
matrix ans;
memset(ans.m, 0, sizeof(ans.m));
for (int i = 0; i < 2; i++) ans.m[i][i] = 1;
while (n) {
if (n & 1) ans = ans * a;
a = a * a;
n >>= 1;
}
return ans;
}
int main() {
ll n;
while (cin >> n ,n != EOF)
{
if (n == -1)
break;
if (n == 0)
{
printf("0\n");
continue;
}
matrix a;
a.m[0][0] = 1; a.m[0][1] = 1;
a.m[1][0] = 1; a.m[1][1] = 0;
a = pow_matrix(a, n);
printf("%lld\n", a.m[0][1]);
}
return 0;
}