HDU1568

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1568

 

package D0720;/* * 思路：要求斐波拉契数列的第n向的前4位数，如果直接按递推公式算肯定超时了 * 通项公式：f(n) = (1/sqrt(5)){[(1+sqrt(5))/2]^n-[(1-sqrt(5))/2]^n} * 由于斐波拉契数列的第20项刚好是4位数，所以前20项可以单独用递推公式计算 * 后面的项可以这样考虑 * 两边取对数 * log10(f(n))=log10(1/sqrt(5))+log10([(1+sqrt(5))/2]^n-[(1-sqrt(5))/2]^n) * 由于[(1-sqrt(5))/2]^n在n》=20时可以忽略不计，所以 * log10(f(n))=log10(1/sqrt(5))+n*log10([(1+sqrt(5))/2]) * f(n)=10^(log10(1/sqrt(5))+n*log10([(1+sqrt(5))/2])) * 由于10的任何次幂都是10的倍数 * 所以f(n)的前4位值只与10^(log10(1/sqrt(5))+n*log10([(1+sqrt(5))/2]))的小数部分有关 * 所以就。。。。开始写代码吧 *  * */import java.io.*;public class HDU1568 {public static void main(String[] args) throws IOException {StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));// 前20项int[] arr = { 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377,610, 987, 1597, 2584, 4181, 6765 };int n;while(st.nextToken()!=StreamTokenizer.TT_EOF){n = (int)st.nval;if(n<=20){System.out.println(arr[n]);continue;}double m = Math.log10(1.0/Math.sqrt(5))+n*Math.log10((1+Math.sqrt(5))/2);double p = m-(long)m;p = Math.pow(10, p);while(p<1000)p*=10;out.println((int)p);}out.flush();}}