C++求1到n中1出现的次数以及数的二进制表示中1的个数

#include "stdio.h"
#include "stdlib.h"
 
int count1(int n);
int count2(int n);
 
int main(void)
{
  int x;
 
  printf("输入一个数：");
  scanf("%d",&x);
  printf("\n从0到%d一共遇到%d（%d）个1\n",x,count1(x),count2(x));
   
  return 0;
}
 
//解法一
int count1(int n)
{
  int count = 0;
  int i,t;
   
  //遍历1到n
  for(i=1;i<=n;i++)
  {
    t=i;
    //依次处理当前遍历到的数字的各个位
    while(t != 0)
    {
        //若为1则统计加一
      count += (t%10 == 1)?1:0;
      t/=10;
    }
  }
   
  return count;
}
 
//解法二：
int count2(int n)
{
  int count = 0;//统计变量
  int factor = 1;//分解因子
  int lower = 0;//当前处理位的所有低位
  int higher = 0;//当前处理位的所有高位
  int curr =0;//当前处理位
   
  while(n/factor != 0)
  {
    lower = n - n/factor*factor;//求得低位
    curr = (n/factor)%10;//求当前位
    higher = n/(factor*10);//求高位
     
    switch(curr)
    {
      case 0:
        count += higher * factor;
        break;
      case 1:
        count += higher * factor + lower + 1;
        break;
      default:
        count += (higher+1)*factor;
    }
     
    factor *= 10;
  }
   
  return count;
}

#include "stdio.h"
#include "stdlib.h"
 
int count1(int x);
int count2(int x);
int count3(int x);
 
int main(void)
{
  int x;
  printf("输入一个数:\n");
  setbuf(stdin,NULL);
  scanf("%d",&x);
  printf("%d转二进制中1的个数是:",x);
  printf("\n解法一：%d",count1(x));
  printf("\n解法二：%d",count2(x));
  printf("\n解法三：%d",count3(x));
   
  printf("\n");
  return 0;
}
 
//除二、余二依次统计每位
int count1(int x)
{
  int c=0;
  while(x)
  {
    if(x%2==1)
      c++;
    x/=2;
  }
  return c;
}
 
//向右移位，与1按位与统计每位
int count2(int x)
{
  int c=0;
  while(x)
  {
    c+=x & 0x1;
    x>>=1;
  }
  return c;
}
 
//每次将最后一个1处理成0，统计处理次数
int count3(int x)
{
  int c=0;
  while(x)
  {
    x&=(x-1);
    c++;
  }
  return c;
}