import java.util.ArrayList; import java.util.List; public class StrTest { public static void main(String[] args) { //模拟生成一个大整数 Long n=ShengCheng(); //拆分这个大整数,看看是由哪些东东组成的 List<Long> list =SplitNumber(n); ;i<list.size();i++) { System.out.println(li
求整数最大的连续0的个数 A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N. For example, number 9 has binary representation 1001 and contains a binary g
Network of Schools Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 13800 Accepted: 5504 Description A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a li
题目描述 输入一个整数,输出该数二进制表示中1的个数.其中负数用补码表示. 直观思路就是把二进制表示从右往左统计1的个数.直接想到移位操作来迭代处理.坑点在于负数的移位操作会填充1.有人贴出了逻辑移位操作,但还是麻烦.直接按照int的位数,32或64,做这么多次移位操作就好了,每次移位操作累计是否为1.int的位数是32还是64,可以写一个函数来做到,而不是硬编码. class Solution { public: int NumberOf1(int n) { int cnt = 0; int
public class Test2 { public static void main(String args[]){ int num; int count[]=new int[21]; for(int i=0;i<10000;i++){ num=(int)(Math.random()*20+0.5); //产生0到20的随机数 count[num]++; //若产生随机数是0,则用count[0]表示它的个数,数组的初始值都为0 System.out.print(num+" "
int getSum(int* arr, int len) { int sum = 0; for (int i = 0; i < len; ++i) { sum += arr[i]; } return sum; } void difPrint(int* arr, int len, vector<int> vct) { cout << vct.size() << " =m 数组: "; for (int i = 0; i < vct.siz
［抄题］: There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinate
package cookie; public class CountBinary_1 { public static void main(String[] args) { System.out.println(count(12)); } // n为自然数 public static int count(int n) { int count = 0; while (n > 0) { count++; n = n & (n - 1); } return count; } } 参考<剑指of
的幂 boolean power2(int x) { return((x&(x-1))==0)&&(x!=0): } For example: #include<stdio.h> int main() { printf(" *******int a=2; int b=3**********\n;"); int a=2; int b=3; printf("计算a&b: %d\n",a&b); printf("
题目链接 Problem Description Galen Marek, codenamed Starkiller, was a male Human apprentice of the Sith Lord Darth Vader. A powerful Force-user who lived during the era of the Galactic Empire, Marek originated from the Wookiee home planet of Kashyyyk as
