http://poj.org/problem?id=3070
根据本题算矩阵,用快速幂即可。
裸题
#include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> using namespace std; #define rep(i, n) for(int i=0; i<(n); ++i) #define for1(i,a,n) for(int i=(a);i<=(n);++i) #define for2(i,a,n) for(int i=(a);i<(n);++i) #define for3(i,a,n) for(int i=(a);i>=(n);--i) #define for4(i,a,n) for(int i=(a);i>(n);--i) #define CC(i,a) memset(i,a,sizeof(i)) #define read(a) a=getint() #define print(a) printf("%d", a) #define dbg(x) cout << #x << " = " << x << endl #define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; } inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; } inline const int max(const int &a, const int &b) { return a>b?a:b; } inline const int min(const int &a, const int &b) { return a<b?a:b; } typedef int matrix[2][2]; matrix a, b; const int M=10000; int n; inline void mul(matrix a, matrix b, matrix c, const int &la, const int &lb, const int &lc, const int &MOD) { matrix t; rep(i, la) rep(j, lc) { t[i][j]=0; rep(k, lb) t[i][j]=(t[i][j]+(a[i][k]*b[k][j])%MOD)%MOD; } rep(i, la) rep(j, lc) c[i][j]=t[i][j]; } int main() { while(~scanf("%d", &n) && n!=-1) { b[0][0]=b[1][1]=1; a[0][0]=a[0][1]=a[1][0]=1; b[0][1]=b[1][0]=a[1][1]=0; while(n) { if(n&1) mul(a, b, b, 2, 2, 2, M); mul(a, a, a, 2, 2, 2, M); n>>=1; } printf("%d\n", b[1][0]); } return 0; }
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