C语言 如何从C中的数字中提取数字？从最高有效数字开始？
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原文地址: http://stackoverflow.com/questions/26688413/
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How to extract digits from a number in C? Begining from the most significant digit?
提问by Brian Brown
Getting digits from a number beginning from the least significant in C is pretty easy:
从 C 中从最不重要的数字开始获取数字非常容易：
#include <stdio.h>
int main()
{
int num = 1024;
while(num != 0)
{
int digit = num % 10;
num = num / 10;
printf("%d\n", digit);
}
return 0;
}
But how to extract digits beginning from the first digit (here 1), that the solution could be applied to any number?
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？
It would be trivial with arrays, but I don't want to use array, and I don't want to use logical operators.
使用数组会很简单，但我不想使用数组，也不想使用逻辑运算符。
回答by Spikatrix
The following program does what you want:
以下程序执行您想要的操作：
#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;
}
回答by rcgldr
Assuming a 32 bit signed number that is positive, then as a simple example:
假设一个 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);
}
or a slight optimization:
或者稍微优化一下：
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);
}
回答by Chandru
Try to store the return value in integer or character array.
尝试将返回值存储在整数或字符数组中。
So I can print the value as we required.
所以我可以根据需要打印值。
If we use character array we can easily find the length and get the result easily.
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]='#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;
}
';
printf("%s",a);
return 0;
}
}
回答by Blackhat002
If you don't want to use array, a simple solution will be..
如果您不想使用数组，一个简单的解决方案将是..
Take Input from the user.
Reverse the number.
And then print the digits.
#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; }
从用户那里获取输入。
倒数。
然后打印数字。
#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; }
回答by Fabulous
I know the question has been answered, but I shall post this answer in case anyone finds it useful.
我知道这个问题已经得到回答，但我会发布这个答案，以防有人觉得它有用。
#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
*/
回答by NeutronStar
Here is my try. Works for positive numbers only. Max range 2^64 (unsigned long long)
这是我的尝试。仅适用于正数。最大范围 2^64 (unsigned long long)
##代码##