生成所有可能的0和& amp;组合的算法1s，对于任何长度的数字 [英] Algorithm to generate all possible combinations of 0s & 1s, for any length of digits

问题描述

我想知道如何打印n个1和0的组合.组合数n是用户定义的.预期的输出是；

I would like to know how can I print n number of combinations of 1s and 0s. The number of combinations, n is user defined. The expected outputs are;

n = 1;

0,1

n = 2;

00,01,10,11

n = 3;

000,001,010,011,100,101,110,111

等.等等.

输出将具有2^n个组合数(其中n是单个组合中的预期位数).

The outputs will have 2^n number of combinations (where n is the number of expected digits in a single combination).

如何在不使用任何内置函数的情况下执行此操作?这个问题与语言无关，是针对算法的.

How can I do this without using any built in function? The question is language independent and is for algorithm.

推荐答案

您可以枚举直到2^n - 1的所有数字.这将使您拥有相同的组合.

You could just enumerate all numbers till 2^n - 1 in binary. That will leave you with the same combination.

n = 2枚举直到2^3 - 1 = 7 转换为二进制:

n = 2 enumerate till 2^3 - 1 = 7 Convert to binary:

000 --> 0
001 --> 1
010 --> 2
011 --> 3
100 --> 4
101 --> 5
110 --> 6
111 --> 7

编辑:也修复了数字位数.可行

EDIT: Fixed the number of digits as well. This works

#include <stdio.h>
#define LENGTH 3
void print_binary(int n)
{
        int bit = 1<<LENGTH - 1;
        while ( bit ) {
        printf("%d", n & bit ? 1 : 0);
        bit >>= 1;
        }
        printf("\n");
}
int main(){
    int n = 1<<LENGTH, i; 
    for(i=0;i<n;i++)
        print_binary(i);
}

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