Fibonacci
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 6131 |
|
Accepted: 4296 |
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:
.
Source
呃，矩阵乘法快速幂嘛
，
很好写
1A
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cassert>
#include <iostream>
#include <sstream>
#include <fstream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#include <iomanip>
using namespace std;
struct node
{
int a,b,c,d;
};
int n;
node mul(node t1,node t2)
{
node tmp;
tmp.a=t1.a*t2.a+t1.b*t2.c;
tmp.b=t1.a*t2.b+t1.b*t2.d;
tmp.c=t1.a*t2.c+t1.c*t2.d;
tmp.d=t1.b*t2.c+t1.d*t2.d;
tmp.a=tmp.a%10000;
tmp.b=tmp.b%10000;
tmp.c=tmp.c%10000;
tmp.d=tmp.d%10000;
return tmp;
}
node get(node a1,int k)
{
node tmp;
if(k==1)
{
return a1;
}
else if(k==2) return mul(a1,a1);
else
{
tmp=get(a1,k/2);
if(k&1)
return mul(mul(tmp,tmp),a1);
else return mul(tmp,tmp);
}
}
int main()
{
node ans,a;
a.a=1;
a.b=1;
a.c=1;
a.d=0;
while(scanf("%d",&n)!=EOF&&n!=-1)
{
if(n==0)
{
printf("0\n");
continue;
}
ans=get(a,n-1);
printf("%d\n",ans.a);
}
return 0;
}