标签：tail break div width sam style identity ros map
http://poj.org/problem?id=3070
|
Language:
Fibonacci
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:
|
：：矩阵快速幂模板题，已经给出了转移矩阵T。
记录下自己写的模板。
学习的博客：https://blog.csdn.net/zhangxiaoduoduo/article/details/81807338
代码：
#include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<sstream> #include<vector> #include<stack> #include<deque> #include<cmath> #include<map> #define sd(x) scanf("%d",&x) #define ms(x) memset(x,0,sizeof x) #define fr(i,n) for(int i=1;i<=n;i++) using namespace std; typedef long long ll; const int mod=10000; const int INF=0x4f4f4f3f; ll t[3][3],tmp[3][3],res[3][3]; ll f[3][3]; void multi(ll (&a)[3][3],ll b[][3]) { ms(tmp); for(int i=0;i<2;i++) for(int k=0;k<2;k++) for(int j=0;j<2;j++) tmp[i][j]+=(a[i][k]*b[k][j])%mod; for(int i=0;i<2;i++) for(int j=0;j<2;j++) a[i][j]=tmp[i][j]%mod; } /* void pr() { for(int i=0;i<2;i++) { for(int j=0;j<2;j++) printf("%d ",t[i][j]); cout<<‘\n‘; } } */ void mp(ll (&t)[3][3],ll n) { ms(res); res[0][0]=res[1][1]=1; while(n) { //pr(); if(n&1) multi(res,t); multi(t,t); n>>=1; } } void init() { ms(f); f[0][0]=1; ms(t); t[0][0]=t[1][0]=t[0][1]=1; } int main() { while(1) { ll n; cin>>n; if(n==-1) break; if(n==0||n==1) { printf("%lld\n",n); continue; } init(); mp(t,n-1); multi(res,f); printf("%lld\n",res[0][0]%mod); } return 0; }
标签：tail break div width sam style identity ros map
原文地址：https://www.cnblogs.com/studyshare777/p/12875408.html
.
.
.