如何在C/C ++中最多计算1000位数字中的位数 [英] How can I count the number of digits in a number up to 1000 digits in C/C++

问题描述

如何计算C或C ++中最多1000个数字的位数

#include <stdio.h>

int main()
{
    int num,counter=0;

    scanf("%d",&num);

    while(num!=0){
        num/=10;
        counter++;
    }

    printf("%d\n",counter);
}

此代码仅适用于最多10位数字-我不知道为什么.

This code works just for numbers up to 10 digits — I don't know why.

推荐答案

由于大多数计算机无法容纳1000个数字的整数，因此您将不得不对输入进行字符串操作或使用 Big数字库.让我们尝试前者.

Since most computers can't hold an integer that is 1000 digits, you will either have to operate on the input as a string or use a Big Number library. Let's try the former.

当将输入视为字符串时，每个数字都是一个介于'0'到'9'(含)范围内的字符.

When treating the input as a string, each digit is a character in the range of '0' to '9', inclusive.

因此，这归结为计数字符:

So, this boils down to counting characters:

std::string text;
cin >> text;
const unsigned int length = text.size();
unsigned int digit_count = 0;
for (i = 0; i < length; ++i)
{
  if (!std::isdigit(text[i]))
  {
    break;
  }
  ++digit_count;
}
cout << "text has " << digit_count << "digits\n";

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