Fibonacci
Time Limit:1000MSMemory Limit:65536KB64bit IO Format:%I64d & %I64u
SubmitPracticePOJ 3070
Appoint description:
Description
In the Fibonacci integer sequence,F0= 0,F1= 1, andFn=Fn 1+Fn 2forn≥ 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 integern, your goal is to compute the last 4 digits ofFn.
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 ofFn. If the last four digits ofFnare all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., printFnmod 10000).
Sample Input
099999999991000000000-1
Sample Output
0346266875
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:
.
题意：题意：求第n个Fibonacci数mod(m)的结果，当n=-1时，break。其中n(where 0 ≤ n ≤ 1,000,000,000) ，，m=10000；
思路：常规方法肯定超时，这道题学会了用矩阵快速幂求斐波那契。如下图：
A = F(n – 1), B = F(N – 2)，这样使构造矩阵。
#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;struct node{int mp[3][3];}init,res;struct node Mult(struct node x,struct node y){struct node tmp;int i,j,k;for(i=0;i<2;i++)for(j=0;j<2;j++){tmp.mp[i][j]=0;for(k=0;k<2;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){int i,j;struct node tmp;for(i=0;i<2;i++)for(j=0;j<2;j++){if(i==j)tmp.mp[i][j]=1;elsetmp.mp[i][j]=0;}while(k){if(k&1) tmp=Mult(tmp,x);x=Mult(x,x);k>>=1;}return tmp;}int main(){int k;while(~scanf("%d",&k)){if(k==-1) break;init.mp[0][0]=1;init.mp[0][1]=1;init.mp[1][0]=1;init.mp[1][1]=0;res=expo(init,k);printf("%d\n",res.mp[0][1]);}return 0;}
用积极的拼搏迎接雨后的彩虹,相信自己
