Fibonacci
Time Limit: 1000 MS Memory Limit: 65536 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
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:
.
#include <iostream>
#include <string.h>
#include <stdio.h> using namespace std;
#define MOD 10000
int n; struct mat
{
int at[][];
} d; mat momu(mat a,mat b)
{
mat c;
memset(c.at,,sizeof(c.at));
for(int i=; i<n; i++)
{
for(int k=; k<n; k++)
{
if(a.at[i][k])
for(int j=; j<n; j++)
{
c.at[i][j]+=a.at[i][k]*b.at[k][j];
if(c.at[i][j]>MOD)
{
c.at[i][j]%=MOD;
}
}
}
}
return c;
} mat expo(mat p,int k)
{
if(k==)
return p;
mat e;
memset(e.at,,sizeof(e.at));
for(int i=; i<n; i++)
e.at[i][i]=;
if(k==)
return e;
while(k)
{
if(k&)
e=momu(p,e);
p=momu(p,p);
k>>=;
}
return e; } int main()
{
n=;
d.at[][]=;
d.at[][]=;
d.at[][]=;
d.at[][]=;
int t;
while(scanf("%d",&t)!=EOF)
{
if(t==-)
break;
mat dd=expo(d,t);
int ans=dd.at[][]%MOD;
printf("%d\n",ans);
}
return ;
}
约等于模板啦啦啦啦~~~~~~~~~~~