B - Fibonacci
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status
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
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 Code
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
using namespace std;
const int inf=0x3f3f3f3f;
const int mod=10000;
int n = 2;
struct node
{
int mp[20][20];
} init,res;
struct node Mult(struct node x, struct node y)// 矩阵相乘
{
struct node tmp;
int i,j,k;
for(i=0; i<n; i++)
{
for(j=0; j<n; j++)
{
tmp.mp[i][j]=0;
for(k=0; k<n; k++)
{
tmp.mp[i][j]=(tmp.mp[i][j]+x.mp[i][k]*y.mp[k][j])%mod;
}
}
}
return tmp;
};
struct node expo(struct node x, int k)//矩阵快速幂
{
struct node tmp;
int i,j;
for(i=0; i<n; i++)
{
for(j=0; j<n; j++)
{
if(i==j)
{
tmp.mp[i][j]=1;
}
else
{
tmp.mp[i][j]=0;
}
}
}
while(k)
{
if(k%2 == 1)
{
tmp=Mult(tmp,x);
}
x=Mult(x,x);
k>>=1;
}
return tmp;
};
int main()
{
int T,i,j,k;
int ans;
while(~scanf("%d",&k))
{
if(k == -1)
{
break;
}
if(k == 0)
{
printf("0\n");
continue;
}
init.mp[0][0] = 1;
init.mp[0][1] = 1;
init.mp[1][0] = 1;
init.mp[1][1] = 0;
res=expo(init,k);
ans = res.mp[0][1];
printf("%d\n",ans);
}
return 0;
}