C + + to find 1 to n 1 occurrences and number of binary representation of 1 of the number of _c language

Number of occurrences of 1 in positive numbers from 1 to n

Topic:

Enter an integer n to find the number of occurrences of 1 in the decimal representation of the n integers from 1 to n.

For example, input 12, from 1 to 12 of these integers contains 1 of the number 1, 10, 1 1 and 12, 11 of which have occurred 5 times

Code implementation (GCC compilation passed):

#include "stdio.h" #include "stdlib.h" int count1 (int n);
 
int count2 (int n);
 
  int main (void) {int x;
  printf ("Enter a number:");
  scanf ("%d", &x);
   
  printf ("\ n from 0 to%d total%d (%d) 1\n", X,count1 (x), Count2 (x));
return 0;
  //Solution one int count1 (int n) {int count = 0;
   
  int i,t;
    Traverse 1 to N for (i=1;i<=n;i++) {t=i;
      To process the individual bit while (T!= 0) of the currently traversed number, add a count of + + (t%10 = 1) for 1? 1:0;
    t/=10;
} return count; }//Solution two: int count2 (int n) {int count = 0;//statistic variable int factor = 1;//factorization factor int lower = 0;//all lower int higher for the current processing bit = 0;//all High level int Curr =0;//current processing bit (n/factor!= 0) {lower = n-n/factor*factor;//to get low Curr = (n/f Actor)%10;//to find the current bit higher = n/(FACTOR*10),//High position switch (curr) {case 0:count + = higher * FAC
        Tor
      Break
        Case 1:count + = higher * factor + lower + 1;
      Break
    Default:count + = (higher+1) *factor;
     
  }  Factor *= 10;
return count;

 }

Analysis:

The first method is to traverse to n from 1, each of which contains the number of "1" add up, better think.

Method Two is more interesting, the core idea is this: statistics on each one may appear 1 times.

Like 123:

Number of digits appearing 1:1,11,13,21,31,...,91,101,111,121

10 digits appearing in 1:10~19,110~119

Hundreds of 1 of digits appear: 100~123

To sum up the rules of each of the 1 appearing, we can get the method two. Its time complexity is O (len), Len is the number length

The number of 1 in the binary representation of integers
Title: Number of 1 in binary representation of integers

Requirements:

Enter an integer that asks how many 1 are in the binary expression of the integer.

For example, enter 10, because the binary representation is 1010, there are two 1, so output 2.

Analysis:

The solution one is the ordinary processing way, through besides two two statistics 1 number;

The solution two is similar to the solution, and is processed by the right displacement, each with the number of 1 bitwise AND statistic 1.

Solution Three is more wonderful, each time the last one of the number processing into 0, statistical processing times, and then statistics 1 of the number of

Code implementation (GCC compilation passed):

#include "stdio.h"
#include "stdlib.h"
 
int count1 (int x);
int count2 (int x);
int count3 (int x);
 
int main (void)
{
  int x;
  printf ("Enter a number: \ n");
  Setbuf (stdin,null);
  scanf ("%d", &x);
  printf ("%d into binary 1 of the number is:", x);
  printf ("\ n Solution One:%d", Count1 (x));
  printf ("\ n Solution II:%d", Count2 (x));
  printf ("\ n Solution three:%d", Count3 (x));
   
  printf ("\ n");
  return 0;
}
 
Every
int count1 (int x)
{
  int c=0 is
  counted in order of two; while (x)
  {
    if (x%2==1)
      C + +;
    x/=2;
  }
  return c;
}
 
Shift right, with 1 bitwise and statistic per bit
int count2 (int x)
{
  int c=0;
  while (x)
  {
    c+=x & 0x1;
    x>>=1;
  }
  return c;
}
 
Each time the last 1 is processed into 0, the statistical processing times
int count3 (int x)
{
  int c=0;
  while (x)
  {
    x&= (x-1);
    C + +;
  }
  return c;
}