TZOJ--1247: Hat's Fibonacci（大數）
1247: Hat's Fibonacci
時間限制(普通/Java):1000MS/10000MS 記憶體限制:32768KByte描述
A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.
F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)
Your task is to take a number as input, and print that Fibonacci number.
輸入
Each line will contain an integers. Process to end of file.
輸出
For each case, output the result in a line.
Note：
No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.
樣例輸入
100
樣例輸出
4203968145672990846840663646
題目來源
題目連結：http://tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=1247
題目大意：F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)，根據這個公式計算F(n)
資料保證了輸出最多不超過2005位，但我自己人懶不想打表估計2005位的需要開多大的陣列，所以用了一個滑動的陣列進行計算
f[i%4] = f[(i-1)%4]+f[(i-2)%4]+f[(i-3)%4]+f[(i-4)%4]
JAVA程式碼：
import java.io.*;
import java.util.*;
import java.math.*;
public class Main
{
public static void main(String args[]) throws Exception
{
Scanner cin=new Scanner(System.in);
BigInteger f[] = new BigInteger[10];
f[3]=f[2]=f[1]=f[0]=BigInteger.ONE;
while(cin.hasNext())
{
int a=cin.nextInt();
f[3]=f[2]=f[1]=f[0]=BigInteger.ONE;
for(int i=4;i<a;i++)
{
f[(i)%4]=f[(i-1)%4].add(f[(i-2)%4].add(f[(i-3)%4].add(f[(i-4)%4])));
}
System.out.println(f[(a-1)%4]);
}
}
}
PYthon程式碼：
while True: try: n = int(input()) f = [1,1,1,1] for i in range(4,n+1) : f[i%4] = f[i%4]+f[(i-1)%4]+f[(i-2)%4]+f[(i-3)%4] print(f[(n-1)%4]) except EOFError: break