大菲波数
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6555 Accepted Submission(s): 2150
Problem Description
Fibonacci数列,定义如下:
f(1)=f(2)=1
f(n)=f(n-1)+f(n-2) n>=3。
计算第n项Fibonacci数值。
f(1)=f(2)=1
f(n)=f(n-1)+f(n-2) n>=3。
计算第n项Fibonacci数值。
Input
输入第一行为一个整数N,接下来N行为整数Pi(1<=Pi<=1000)。
Output
输出为N行,每行为对应的f(Pi)。
Sample Input
5 1 2 3 4 5
Sample Output
1 1 2 3 5
Source
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lcy
import java.util.*; import java.math.*; public class Main { static BigInteger a[] = new BigInteger[1005]; private static void Fibo() { a[1] = BigInteger.ONE; a[2] = BigInteger.ONE; for (int i = 3; i <= 1000; i++) a[i] = a[i - 1].add(a[i - 2]); } public static void main(String[] args) { Fibo(); Scanner in = new Scanner(System.in); int t = in.nextInt(); for (int i = 0; i < t; i++) { int n = in.nextInt(); System.out.println(a[n]); } } }