如果先不考慮 60 個 LED 的問題，256 可以亮 9 個燈，我想到 2 種做法給您參考。

陣列版本

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int input = 256;

    int led[9];

    for(int i=0; i<9; i++)
    {
        led[i] = (input & 1 << i) != 0;
    }

    for(int i=0; i<9; i++)
    {
        printf("%d", led[i]);
    }
    printf("\n");

    return 0;
}

測試連結

不能用陣列版本

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int input = 256;

    int led1, led2, led3, led4, led5, led6, led7, led8, led9;
    
    led1 = (input & 1) != 0;
    led2 = (input & 2) != 0;
    led3 = (input & 4) != 0;
    led4 = (input & 8) != 0;
    led5 = (input & 16) != 0;
    led6 = (input & 32) != 0;
    led7 = (input & 64) != 0;
    led8 = (input & 128) != 0;
    led9 = (input & 256) != 0;

    printf("%d", led1);
    printf("%d", led2);
    printf("%d", led3);
    printf("%d", led4);
    printf("%d", led5);
    printf("%d", led6);
    printf("%d", led7);
    printf("%d", led8);
    printf("%d", led9);
    printf("\n");

    return 0;
}

測試連結

C++輸出寫法，value為輸入值。

#include <iostream>
int main()
{
    long long value = 4;
    
    long long nowValue = 1;
    for(unsigned int index = 0; index < 60; index++)
    {
        if((nowValue & value) == nowValue)
        {
            std::cout << index + 1 <<"開" << std::endl;
        }
        nowValue <<= 1;
    }

    return 0;
}