#include<iostream> using namespace std; int main() { long long start, end , i, check, b, c, cnt=; cin >> start >> end >> check;//输入范围与要查的数字: for (i = start; i <= end; i++)//一到n进行循环: { b = i;//为了不改变i的值,就把i赋值给一个数: while (b != )//如
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example:Given n = 13,Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13. Hint: Beware of overflow. c
个位ch个数 + 十位ch个数 * 10 + 百位ch个数 * 100:同时如果某一位刚好等于ch,还需要减去多算的一部分值. #include <stdio.h> //整数1到n,字符ch出现的次数:如1到12,1出现5次 int count1s (int n, char ch){ int i = ch - '0'; int count = 0; int j = 1; int m = n; int t = 0; int k = 1; if (i == 0) i = 10; //循环比较最后一
计算一个无符整数中1Bit的个数(1) 2010-04-20 10:52:48 分类: C/C++ [转]计算一个无符整数中1Bit的个数(1) Count the number of bits that are on in an unsigned integer(计算一个无符整数中1Bit的个数)-- (1) 计算一个无符号整数中有多少的Bit为1 这是一个经常遇到的经典问题,这里分两个部分讲解和总结,首先对讲解现有的算法,然后再讲解一些改进算法. 1.循环法(Iterated Count
敌兵布阵 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 59891 Accepted Submission(s): 25331 Problem Description C国的死对头A国这段时间正在进行军事演习,所以C国间谍头子Derek和他手下Tidy又开始忙乎了.A国在海岸线沿直线布置了N个工兵营 地,Derek和Tidy的任务