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).
题解：
矩阵快速幂与斐波那契数列的基础题。
题目公式已经给过了，直接套模板就行了。
如果不了解矩阵快速幂，点这里小哥哥小姐姐~(^_^)~
代码：
#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
typedef long long LL;
const int mod=1e4;
struct mat
{
LL a[2][2];
};
LL n;
mat tc(mat x,mat y)
{
mat res;
memset(res.a,0,sizeof(res.a));
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
for(int k=0;k<2;k++)
{
res.a[i][j]+=x.a[i][k]*y.a[k][j];
res.a[i][j]%=mod;
}
return res;
}
LL pow_mod()
{
mat res,c;
c.a[0][0]=1; c.a[0][1]=1;
c.a[1][0]=1; c.a[1][1]=0;
memset(res.a,0,sizeof(res.a));
for(int i=0;i<2;i++)
res.a[i][i]=1;
while(n>0)
{
if(n&1) res=tc(res,c);
c=tc(c,c);
n>>=1;
}
return res.a[0][1];
}
int main()
{
while(scanf("%I64d",&n))
{
if(n==-1)
break;
LL d=pow_mod()%mod;
if(d==0000)
printf("0\n");
else
printf("%I64d\n",d);
}
return 0;
}