How many & #39; 1 & #39; s are there and there

How many '1' s are there solutions
Description:

Description:

Input number n (n <= 50) in the first line, indicating that there are n groups of test cases. input number m (m is an integer) in each line from row 2nd to row n + 1 ), count and output the number of 1 in the binary representation of m.

For example, if m = 9, if the binary value is 1001, 2 is output.

Input:

2

3

7

Output:

2

3

 

 

Hint:

Bitwise operations

A relatively simple question, which can also have advanced practices. Although bit operations are prompted, bit operations are not used because you are not familiar with bit operations. Wrong code: (the computer manager directly regards it as a spyware and killed it ..) If you do not understand the problem, post it first.

#include <stdio.h>int main() {    int binary[20];    int count = 0;    int one = 0;      int n, i, num;    scanf("%d", &n);    for (i = 0; i < n; i++) {        count = 0;        one = 0;        scanf("%d", &num);        while (num != 0) {                 binary[count] = num % 2;               num /= 2;            count++;        }        for (i = count-1; i >= 0; i--) {               if (binary[i] == 1)            one++;          }        printf("%d\n", one);    }}

Code passed:

#include <stdio.h>int main() {    int binary[20];    int n, i, num, j;    scanf("%d", &n);    for (i = 0; i < n; i++) {        int count = 0;        int one = 0;        scanf("%d", &num);        while (num != 0) {            binary[count] = num % 2;             if (binary[count] == 1) {                one++;            }            num /= 2;            count++;        }        printf("%d\n", one);    }}

Answer:

#include<stdio.h> int bitcount(int x) {        int count = 0;        while (x != 0) {                x &= (x-1);                count++;        }        return count;} int main() {        int num;        int x;        scanf("%d", &num);         while (num--) {                scanf("%d", &x);                printf("%d\n", bitcount(x));        }        return 0;}

 

I heard there is a function.