分析一波
典型的斐波那契數(shù)列應(yīng)用。
分析：當(dāng) n = 1 時瘦癌，只有一種跳法誓沸；當(dāng) n = 2 時否副，有兩種；
當(dāng) n > 2 時愕乎，
如果第一次跳 1 級阵苇，則跳法總數(shù) = F(n-1)：后面剩下的 n - 1 級臺階的跳法總數(shù)；
如果第一次跳 2 級感论，則跳法總數(shù) = F(n-2)：后面剩下的 n - 2 級臺階的跳法總數(shù)绅项；因此 n 級臺階的不同跳法的總數(shù)：F(n) = F(n-1) + F(n-2)。
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
/**
* 題意：
*/
public class Main {
private static int arr[] = new int[33];
public static void FibonacciPlus() {
arr[1] = 1;
arr[2] = 2;
for (int i = 3; i <= arr.length; i++) {
arr[i] = arr[i - 1] + arr[i - 2];
}
}
public static void main(String[] args) throws IOException {
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
int n;
FibonacciPlus();
while (in.nextToken() != StreamTokenizer.TT_EOF) {
n = (int) in.nval;
out.println(arr[n]);
}
out.flush();
}
}
