标签：des style blog http color io os java ar
Time Limit: 1000 MS Memory Limit: 65536 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
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.
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.
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).
0 9 999999999 1000000000 -1
0 34 626 6875
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:
.
#include <iostream> #include <string.h> #include <stdio.h> using namespace std; #define MOD 10000 int n; struct mat { int at[2][2]; } d; mat momu(mat a,mat b) { mat c; memset(c.at,0,sizeof(c.at)); for(int i=0; i<n; i++) { for(int k=0; k<n; k++) { if(a.at[i][k]) for(int j=0; j<n; j++) { c.at[i][j]+=a.at[i][k]*b.at[k][j]; if(c.at[i][j]>MOD) { c.at[i][j]%=MOD; } } } } return c; } mat expo(mat p,int k) { if(k==1) return p; mat e; memset(e.at,0,sizeof(e.at)); for(int i=0; i<n; i++) e.at[i][i]=1; if(k==0) return e; while(k) { if(k&1) e=momu(p,e); p=momu(p,p); k>>=1; } return e; } int main() { n=2; d.at[0][0]=1; d.at[0][1]=1; d.at[1][0]=1; d.at[1][1]=0; int t; while(scanf("%d",&t)!=EOF) { if(t==-1) break; mat dd=expo(d,t); int ans=dd.at[0][1]%MOD; printf("%d\n",ans); } return 0; }
约等于模板啦啦啦啦~~~~~~~~~~~
标签：des style blog http color io os java ar
原文地址：http://www.cnblogs.com/zhangying/p/3975699.html
