POJ-3070 Fibonacci (矩阵快速幂)
Time Limit: 1000MS | Memory Limit: 65536K | |
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
这道题初始值是1,少了些步骤。简化了
F(n)=f(n-1)+f(n-2)
列矩阵的元素个数是由n-x的x决定的
#include<cstdio> using namespace std; int n,base[2][2],ans[2][2]; void mul(int b[2][2],int a[2][2],int tans[2][2]) { int i,j,k; int t[2][2]; for(i=0;i<2;i++) for(j=0;j<2;j++) { t[i][j]=0; for(k=0;k<2;k++) t[i][j]=(t[i][j]+b[i][k]*a[k][j])%10000; //在答案出取余就好 } for(i=0;i<2;i++) for(j=0;j<2;j++) tans[i][j]=t[i][j]; } void pow_mod() //base[][],ans[][],n定成全局变量了 { base[0][0]=base[0][1]=base[1][0]=1;base[1][1]=0; ans[0][0]=ans[1][1]=1; ans[1][0]=ans[0][1]=0; while(n) { if((n&1)==1) mul(base,ans,ans); mul(base,base,base); n>>=1; } } int main() { while(scanf("%d",&n)) { if(n==-1) break; pow_mod(); printf("%d\n",ans[1][0]); } return 0; }