Fibonacci
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 12951 Accepted: 9210
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
Stanford Local 2006
入门题目
代码
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int MAXN =100+10;
const int mod=10000;
struct Matrix{
LL a[MAXN][MAXN];
int r,c;
};
Matrix ori,res;
void init(){
memset(res.a,0,sizeof(res.a));
res.r=2;res.c=2;
for(int i=1;i<=2;i++) //将结果矩阵赋值为单位矩阵
res.a[i][i]=1;//单位矩阵A就是 [未知矩阵]B*A=B;
ori.r=ori.c=2;
ori.a[1][1]=ori.a[1][2]=ori.a[2][1]=1;// 初始化初始矩阵
ori.a[2][2]=0;
}
Matrix multi(Matrix x,Matrix y){
Matrix z;
memset(z.a,0,sizeof(z.a));
z.r=x.r; z.c=y.c; //新矩阵的行数等于x的行数,列数等于y的列数
for(int i=1;i<=x.r;i++){
for(int k=1;k<=x.c;k++){
if(x.a[i][k]==0) continue;
for(int j=1;j<=y.c;j++)
z.a[i][j]=(z.a[i][j]+(x.a[i][k]*y.a[k][j]%mod))%mod;
}
}
return z;
}
void Matrix_mod(int n){
while(n){
if(n&1) res=multi(ori,res);
ori=multi(ori,ori);
n>>=1;
}
printf("%lld\n",res.a[1][2]%mod);
}
int main(){
int n;
while(scanf("%d",&n)&&(n!=-1)){
init();
Matrix_mod(n);
}
return 0;
}