Fibonacci
Time Limit: 1000MS | | Memory Limit: 65536K |
Total Submissions: 13219 | | Accepted: 9394 |
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
09 999999999 1000000000 -1
Sample Output
034 626 6875
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#define maxn 1000005
#define MOD 10000
using namespace std;
typedef long long ll;
int p1[2][2], p2[2][2], p3[2][2];
void Multi(int (*k1)[2], int (*k2)[2], int (*k3)[2]){
for(int i = 0; i < 2; i++)
for(int j = 0; j < 2; j++){
k3[i][j] = k1[i][0] * k2[0][j] + k1[i][1] * k2[1][j];
k3[i][j] %= MOD;
}
}
void solve(int n){
while(n){
if(n&1){
Multi(p1, p2, p3);
memcpy(p2, p3, sizeof(p3));
}
Multi(p1, p1, p3);
memcpy(p1, p3, sizeof(p3));
n >>= 1;
}
}
int main(){
//freopen("in.txt", "r", stdin);
int n;
while(scanf("%d", &n) == 1 && n != -1){
if(n == 0)
puts("0");
else if(n == 1)
puts("1");
else{
p2[0][0] = 1, p2[0][1] = 0;
p2[1][0] = 0, p2[1][1] = 1;
p1[0][0] = 1, p1[0][1] = 1;
p1[1][0] = 1, p1[1][1] = 0;
solve(n-1);
printf("%d\n", p2[0][0]);
}
}
return 0;
}
