Question

从C中从最不重要开始的数字中获取数字非常简单：

#include <stdio.h>

int main()
{
    int num = 1024;

    while(num != 0)
    {
        int digit = num % 10;
        num = num / 10;
        printf("%d\n", digit);
    }


    return 0;
}

但是如何从第一个数字（此处为1）开始提取数字，该解决方案可以应用于任何数字？

使用数组会很简单，但我不想使用数组，我也不想使用逻辑运算符。

Answer 1

以下程序可以满足您的需求：

#include <stdio.h>

int main()
{

    int num =0;
    int power=1;

    printf("Enter any number:");
    scanf("%d",&num);

    while(num>power)
      power*=10;

    power/=10;

    while(num != 0)
    {
        int digit = num /power;
        printf("%d\n", digit);
        if(digit!=0)
          num=num-digit*power;
        if(power!=1)
          power/=10;
    }



    return 0;
}

Answer 2

假设32位有符号数是正数，那么作为一个简单的例子：

#include <stdio.h>

int main()
{
int rmdr;
int dvsr;
int quot;
    scanf("%d", &rmdr);               // read in number
    dvsr = 1000000000;
    while(0 >= (quot = rmdr / dvsr)){ // skip leading zeroes
        rmdr %= dvsr;
        dvsr /= 10;
        if(dvsr == 1)
            break;
    }
    while(dvsr){                      // display number
        quot = rmdr / dvsr;
        printf("%1d", quot);
        rmdr %= dvsr;
        dvsr /= 10;
    }
    printf("\n");
    return(0);
}

或略微优化：

int main()
{
int rmdr;
int rm10;
int dvsr;
int quot;
    scanf("%d", &rmdr);               // read in number
    rm10 = rmdr/10;
    dvsr = 1;
    while(dvsr <= rm10)               // skip to 1st digit
        dvsr *= 10;
    while(dvsr){                      // display number
        quot = rmdr / dvsr;
        printf("%1d", quot);
        rmdr %= dvsr;
        dvsr /= 10;
    }
    printf("\n");
    return(0);
}

Answer 3

尝试将返回值存储为整数或字符数组。

所以我可以按照我们的要求打印这个值。

如果我们使用字符数组，我们可以轻松找到长度并轻松获得结果。

#include <stdio.h>
#include<string.h>
int main()
{
    int num = 1024;
char a[10];
        int i=0;
    while(num != 0)
    {
        a[i++] = (num % 10)+'0';
        num = num / 10;

    }
    a[i]='\0';
    printf("%s",a);
    return 0;

}

Answer 4

如果您不想使用数组，一个简单的解决方案就是......

从用户那里获取输入。
撤消号码。

然后打印数字。

#include<stdio.h>
#include <math.h>

int reversDigits(int num)
{
int rev_num = 0;
while(num > 0)
{
rev_num = rev_num*10 + num%10;
num = num/10;
}
return rev_num;
}

int main() {
int i = 1024;
int number = reversDigits(i);
while(number != 0)
{
    int digit = number % 10;
    number = number/ 10;
    printf("%d\n", digit);
}
return 0;
}

Answer 5

我知道这个问题已得到解答，但我会发布这个答案，以防有人发现它有用。

#include<stdio.h>

//get_int_first gets each digit and multiplies it by it's placeholder value.
// the number is reconstructed and returned to main

int get_int_first(int num){
    int power = 1,len = 0,first = 0,i = 0;
    int number = 0;

    while (num>power){                                          
        power=power*10;                                 //calculating number of zeroes in number. for 789, power = 10 -> 100 -> 1000
    }
    power = power/10;                                   // to multiply with power directly and get highest placeholder, we divide by 10. now power = 100

    while (power>1){                                    
        first = (num/power);                            //get digits from the leftmost(100th placeholder/nth placeholder)  f = 7                    f = 8
        number = number + first*power;                  //first*power = gives number as                                    nb = 0 +7*100 = 700      nb = 700 + 80
        num = num - (power*(first));                    //change number to get next significant digit from left            n = 789 - (100*7) = 89   n = 89-10*8=9
        power = power/10;                               //change power to divide number by appropriate power.              p = 100/10 = 10          p = 1
    }
    number = number + num;                              //number is missing the unit's digit and it is stored in num. thus, it is added to number
    return number;
}


int main() {

    printf("digits are %d\n",get_int_first(789));
    return 0;
}

Answer 6

这是我的尝试。仅适用于正数。最大范围2 ^ 64（unsigned long long）

#include <iostream>
#include <cmath>

using namespace std;
using bignum = unsigned long long;

inline
bignum Power(unsigned x, unsigned y) {
    return y>0 ? x*Power(x,y-1) : 1;
}

// return digits count in a number
inline
int Numlen(bignum num) {
    return num<10 ? 1 : floor(log10(num))+1;
}

// get the starting divisor for our calculation
inline
bignum Getdivisor(unsigned factor) {
    return Power(10, factor);
}


int main()
{
    bignum num{3252198};
    //cin >> num;

    unsigned numdigits = Numlen(num);    
    auto divisor = Getdivisor(numdigits-1);

    while(num > 0) {

        cout << "digit = " << (num/divisor) << "\n";

        num %= divisor;
        divisor /= 10;
    }
}

/*
output:
digit = 3
digit = 2
digit = 5
digit = 2
digit = 1
digit = 9
digit = 8
*/

如何从C中的数字中提取数字？从最重要的数字开始？

6 个答案: