时间限制: 1 Sec 内存限制: 128 MB
Yang Hui’s Triangle is defined as follow. In the first layer, there are two numbers A1,1 and A1,2 satisfying A1,1 = A1,2 = 1. Then for each i > 1, the i-th layer contains i + 1 numbers satisfying Ai,1 = Ai,i+1 = 1 and Ai,j = Ai−1,j−1 + Ai−1,j for 1 < j ≤ i.1 1
1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1Now, given an integer N , you are asked to count the number of even integers in the first N layers.
输入 The input file contains multiple cases, please handle it to the end of file. For each case, there is only one line containing an integer N (0 < N ≤ 1050). 输出 For each case, output the number of the even integers in the first N layers of Yang Hui’s Triangle. 样例输入 4 8 12 样例输出 4 16 42
求杨辉三角前n行偶数个数和
先打了个表
1 1 1 //题目中 N=1 从这行开始，但为了便于发现规律，补上了第一行 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 0 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 1 0 0 0 1 0 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 1 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 1 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1统计奇数的个数，取补集
统计奇数的个数时： 令C = N,base=1，重复如下步骤直到C等于零: 1.找到一个最大的k,使得2k<=C 2.C-=2k 3.答案统计ans+=3k(由表中规律得)*base 4.base*=2(递归分形规律)import java.util.*;import java.math.*;public class Main { static Scanner cin = new Scanner(System.in); public static void main(String[] args) { while(cin.hasNextBigInteger()){ BigInteger n = cin.nextBigInteger(); n = n.add(BigInteger.ONE); BigInteger cnt = BigInteger.ONE; BigInteger ans = BigInteger.ZERO; BigInteger sq = (n.add(BigInteger.ONE).multiply(n).divide(BigInteger.valueOf(2))); while (n.compareTo(BigInteger.valueOf(0)) > 0) { for (int i = 170; i >= 0; i--) { BigInteger m2 = BigInteger.valueOf(2).pow(i); if (n.compareTo(m2) >= 0) { n = n.subtract(m2); BigInteger m3 = BigInteger.valueOf(3).pow(i); ans = ans.add(m3.multiply(cnt)); cnt = cnt.multiply(BigInteger.valueOf(2)); } } } System.out.println(sq.subtract(ans)); } }}