How many '1's are there題解，there題解

How many '1's are there題解，there題解

Description:

Description:

第一行輸入數字n（n<=50)，表示有n組測試用例，第2到第n+1行每行輸入數m（m為整數），統計並輸出m用二進制表示時，1的個數。

例如：m=9時，二進制表示為1001，則輸出2.

Input:

2

3

7

Output:

2

3

Hint:

利用位運算

#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);
    }
}

最後通過的代碼：

#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);
    }
}

標答：

#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;
}

聽說有一個函數