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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10931 | Accepted: 7770 |
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
就是矩阵快速幂，裸题。
#include <stdio.h>
#include <math.h>
#include <vector>
#include <queue>
#include <string>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define Mod 10000
using namespace std;
int res[5][5];
int mat[5][5];
void Matmul(int x[5][5],int y[5][5])
{
int t[5][5]={0};
for(int i=0;i<2;i++){
for(int k=0;k<2;k++){
if(x[i][k]){
for(int j=0;j<2;j++)
t[i][j]=(t[i][j]+(x[i][k]*y[k][j]))%Mod;
}
}
}
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
x[i][j]=t[i][j];
}
}
}
void Matrix(int x[5][5],int n)
{
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
res[i][j]=(i==j);
}
}
while(n){
if(n&1)Matmul(res,x);
Matmul(x,x);
n>>=1;
}
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF){
mat[0][0]=mat[0][1]=mat[1][0]=1;
mat[1][1]=0;
if(n==-1)break;
if(n==0){
printf("0\n");
continue;
}
Matrix(mat,n);
printf("%d\n",res[0][1]);
}
return 0;
}
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原文地址：http://blog.csdn.net/ydd97/article/details/47355373
