Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 11236 | Accepted: 7991 |
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:
.
代码:
#include<stdio.h> #include<string.h> #define MOD 10000 struct node { int m[2][2]; }a, b; node cheng(node x, node y) { int i, j, k; node c; for(i=0; i<2; i++) for(j=0; j<2; j++) { c.m[i][j] = 0; for(k=0; k<2; k++) c.m[i][j] = (c.m[i][j] + x.m[i][k]*y.m[k][j])%MOD; } return c; } int Fast_MOD(int n) { a.m[0][0] = a.m[0][1] = a.m[1][0] = 1; a.m[1][1] = 0; b.m[0][0] = b.m[1][1] = 1; /// b 初始化为单位矩阵 b.m[0][1] = b.m[1][0] = 0; while(n) { if(n&1) /// n是奇数 b = cheng(b, a); a = cheng(a, a); n >>= 1; } return b.m[0][1]; } int main() { int n; while(scanf("%d", &n), n!=-1) { printf("%d ", Fast_MOD(n)); } return 0; }