in python, when uses print, how to control how many digits that a printed number show

You can use string formatting to control the number of digits that are displayed when you print a number in Python. One common way to do this is to use the `format()` method on a string. Here's an example:

x = 3.14159
print("The value of x is {:.2f}".format(x))

In this example, `"{:.2f}"` is a format string that tells Python to display `x` as a floating-point number with 2 decimal places. The `f` in `"{:.2f}"` specifies that the number should be formatted as a float.

Output:

The value of x is 3.14

You can replace the `2` in `"{:.2f}"` with any number to display a different number of decimal places.

相关问题

Write a program in C to count the digits of a given number using recursion. Input: Input a non-negtive integer n. Output: Print the number of digits in the number n. Input Sample: 50 Output Sample: The number of digits in 50 is: 2

Sure, here's a program in C that counts the digits of a given number using recursion:

#include <stdio.h>

// Function to count the number of digits in a number using recursion
int countDigits(int n) {
    if (n == 0) {
        return 0;
    } else {
        return 1 + countDigits(n / 10);
    }
}

int main() {
    int n;

    // Read input from user
    printf("Enter a non-negative integer: ");
    scanf("%d", &n);

    // Count the number of digits using recursion
    int numDigits = countDigits(n);

    // Print the result
    printf("The number of digits in %d is: %d\n", n, numDigits);

    return 0;
}

The `countDigits()` function takes an integer `n` as input and recursively counts the number of digits in `n` by dividing it by 10 until `n` becomes 0. The base case is when `n` is 0, in which case the function returns 0. Otherwise, the function returns 1 plus the result of calling itself with `n/10`.

In the `main()` function, we read the input from the user, call the `countDigits()` function to count the number of digits, and print the result.

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again! Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number. Input Specification: Each input contains one test case. Each case contains one positive integer with no more than 19 digits. Output Specification: For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number. Sample Input: 1234567899 Sample Output: Yes 2469135798

首先，我们需要将输入的数转换成字符串，并计算出它的长度。接着，我们将该数乘以2，再将结果转换成字符串，并且将它的长度与原始字符串的长度进行比较。如果长度不同，那么乘以2得到的数字肯定不是原始数字的排列。如果长度相同，那么我们需要逐一检查乘以2得到的数字中的每一个数字，确保它们在原始数字中都存在且出现的次数相同。如果满足这个条件，那么乘以2得到的数字就是原始数字的排列，否则就不是。

下面是实现这个算法的代码：