Question

我希望能发挥一种激动作用,如果这种愤怒超过某种价值(就我目前的情况而言是1 000人),那么我就希望对其作出某些区分,以便我最终能够恢复原始愤怒的缩略语。

For example:

1000 = 1
1001 = 1
1099 = 1
1100 = 1.1
1199 = 1.1
1200 = 1.2
10000 = 10
10099 = 10
10100 = 10.1

It s the division and rounding side of things that has been causing me problems.

给我带来上述结果的最适当方法是什么?

Answer 1

如何:

int dividedBy100 = x / 100; // Deliberately an integer division
decimal dividedBy1000 = dividedBy100 / 10m; // Decimal division
string result = dividedBy1000.ToString();

我将使用<条码>标准<>。页: 1

Answer 2

缩略语按定义四舍五入。

如果你想更精确地说明为什么不使用双倍而不是惯性?

Answer 3

这里的建议

double function(int number)
{
    if (number >= 1000)
    {
       return (((double)number) / 1000);
    }

}

Answer 4

你的事例似乎暗示,你只想有一个精度的模棱两可的地方,这样就怎么说:

Divide by 100
Cast to double (or float)
Divide by 10

第一个司将缩小任何低于100号的拖车数量(相当于100个基底功能),然后将翻一番,然后将10个线分开,这将给你希望的单一精准地点。

Answer 5

如果是原始编号,则

int a=t/100

float b=a/10

b 包括答复

Some more code:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            while (true)
            {
                string s;
                s = Console.ReadLine();
                int a = Convert.ToInt32(s);
                a = a / 100;
                float b = a / (float)10.0;
                Console.WriteLine(b);
            }
        }
    }
}

Answer 6

You should use modular (remainder) mathematics to do this. You don t need to involve the FPU (Floating Point Unit).

static string RoundAndToString(int value, int denominator)
{
    var remainder = value % denominator;
    value = (value - remainder) / denominator;
    if (remainder == 0)
        return value.ToString();
    remainder = (remainder * 10) / denominator;
    return string.Format("{0}{1}{2}", value, CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator, remainder);
}

Answer 7

既然你只是想缩减人数,那么,把它改为扼杀,从扼杀中删除最后两个特征,然后把10个字分开,以获得相应的数字。

这里是Ruby的算法。 (无C# 手提)

a = 1000 #sample number
 -> 1000

b = a.to_s[0..-3] #a converted to a string, then taking all characters except the last two.
 -> "10"

c = b.to_i / 10.0 # converts to float in correct power
 -> 1.0

You then display "c" in whatever format you want using sprintf (or the C# equivalent using FormatNumber).

Answer 8

审判

int MyNumber = 10100;

string MyString = ((int) MyNumber/1000).ToString() + (( MyNumber % 1000) > 99 ? "." + (((int)( MyNumber / 100 )) % 10).ToString() : "");

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