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
#include<cstdio>
#include<cstring>
using namespace std;
typedef struct Matrix
{
int s[2][2];
} M;
M Multiply(M a,M b)
{
M c;
memset(c.s,0,sizeof(c.s));
for(int i=0;i<2;i++)
{
for(int j=0;j<2;j++)
{
for(int k=0;k<2;k++)
{
c.s[i][j]+=a.s[i][k]*b.s[k][j]%10000;
}
}
}
return c;
}
M pow(M a,int n)
{
M b;
if(n==0)
{
memset(b.s,0,sizeof(b.s));
b.s[0][0]=1;
b.s[1][1]=1;
return b;
}
else
{
M k=pow(a,n/2);
if(n%2==1)
return Multiply(Multiply(k,k),a);
else return Multiply(k,k);
}
}
int main()
{
int n;
while(scanf("%d",&n),n!=-1)
{
M a;
a.s[0][0]=1;
a.s[0][1]=1;
a.s[1][0]=1;
a.s[1][1]=0;
M b=pow(a,n);
printf("%d\n",b.s[0][1]%10000);
}
return 0;
}
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