分配用户输入到预定义的数字中

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英文:

Distribute a user input among the predefined numbers

问题

我正在尝试在Java中构建一个简单的输入/输出程序。该程序应该从用户那里获取输入,然后将其分配给预定义的数字。例如,假设用户输入452,预定义的值是300、200、100、50。程序应该以字符串格式输出如下: "1x300,0x200,1x100,1x50,余数为2。" 但是,我无法弄清楚代码。

英文:

I am trying to build a simple input/output program in java. This program should take an input from a user and then distribute it among the predefined numbers. For example, let's say a user enters 452, and the predefined values are 300, 200, 100, 50. The program should give the output in a string format as follows: "1x300, 0x200, 1x100, 1x50, with a remainder of 2." However, I can't figure out the code.

答案1

得分: 2

这个解决方案应该是有效的:

int[] values = {300, 200, 100, 50};
int[] multipliers = new int[values.length];

int input = 452;

for(int i = 0; i < values.length; ++i)
{
    if(input < values[i])
    {
        continue;
    }
    else
    {
        multipliers[i] = input / values[i];
        input = input % values[i];
        // 因为隐式转换,它会自动变成整数
    }
}

for(int i = 0; i < multipliers.length; ++i)
{
    System.out.println(multipliers[i]+"x"+values[i]);
}

System.out.println("剩余为:"+input);

# 解释

假设值是按降序输入的我们遍历 `values` 数组将每个乘数的元素设置为 `values[i]` 可以整除输入的次数然后我们使用取模(`%`)运算符将输入设置为 `values[i]` 的余数

最后所有的乘数将被正确设置输入将等于余数

<details>
<summary>英文:</summary>

This solution should work:

    int[] values = {300, 200, 100, 50};
    int[] multipliers = new int[values.length];

    int input = 452;

    for(int i = 0; i &lt; values.length; ++i)
    {
        if(input &lt; values[i])
        {
            continue;
        }
        else
        {
            multipliers[i] = input / values[i];
            input = input % values[i];
            // It will automatically be an integer because of implict conversion
        }
    }

    for(int i = 0; i &lt; multipliers.length; ++i)
    {
        System.out.println(multipliers[i]+&quot;x&quot;+values[i]);
    }

    System.out.println(&quot;Remainder is &quot;+input);

# Explanation:

Assuming the values are inputted in descending order, we loop through `values` and set each element of multiplier equal to the number of times `values[i]` can divide the input. Then we set input to the remainder (using the modulo `%` operator) from `values[i]`.

At the end, all the multipliers will be set correctly, and input will be equal to the remainder

</details>



# 答案2
**得分**: 1

```java
public static void main(String[] args){

	//应按降序排列
	//键表示预定义值
	//值表示预定义值的计数
	Map<Integer, Integer> predefinedValues = new LinkedHashMap<>();
	predefinedValues.put(300, 0);
	predefinedValues.put(200, 0);
	predefinedValues.put(100, 0);
	predefinedValues.put(50, 0);

	//获取最小的预定义值
	int lowestPredef = 50;

	//获取输入
	int input = 655;

	//在输入大于最小值的情况下进行迭代
	while(input > lowestPredef){

		//迭代所有预定义值
		for(Map.Entry<Integer, Integer> entry : predefinedValues.entrySet()){

			int key = entry.getKey();
			int value = entry.getValue();

			//找到并在第一个低于输入的值处停止
			if(input > key){

				//从输入中减去该值
				input -= key;

				//该值的计数加1
				predefinedValues.put(key, value + 1);

				//跳出循环
				break;

			}

		}

	}

	//使用结果构建字符串
	StringBuilder result = new StringBuilder("");
	for(Map.Entry<Integer, Integer> entry : predefinedValues.entrySet()){

		result.append(entry.getValue())
				.append("x")
				.append(entry.getKey())
				.append(", ");

	}
	result.append("余数为")
			.append(input);

	System.out.print(result.toString());

}
英文:

Normally, you should provide us the code you have tried, to get help.

So I was free to think the solution in my way.

public static void main(String[] args){
//should be in descending order
//key represents predef value
//value represents the count of predef value
Map&lt;Integer, Integer&gt; predefinedValues = new LinkedHashMap&lt;&gt;();
predefinedValues.put(300, 0);
predefinedValues.put(200, 0);
predefinedValues.put(100, 0);
predefinedValues.put(50, 0);
//get lowestPredef
int lowestPredef = 50;
//get input
int input = 655;
//iterate until input is higher than lowest
while(input &gt; lowestPredef){
//iterate all predefined values
for(Map.Entry&lt;Integer, Integer&gt; entry : predefinedValues.entrySet()){
int key = entry.getKey();
int value = entry.getValue();
//find and stop at first lower value than input
if(input &gt; key){
//substract that value from input
input -= key;
//count +1 for that value
predefinedValues.put(key, value + 1);
//break for-loop
break;
}
}
}
//build a string with result
StringBuilder result = new StringBuilder(&quot;&quot;);
for(Map.Entry&lt;Integer, Integer&gt; entry : predefinedValues.entrySet()){
result.append(entry.getValue())
.append(&quot;x&quot;)
.append(entry.getKey())
.append(&quot;, &quot;);
}
result.append(&quot;with a remainder of &quot;)
.append(input);
System.out.print(result.toString());
}

答案3

得分: 0

这段代码适用于任何数字。

public static void main(String[] args) {
    int value = 452;
    Map<Integer, Integer> listofmultiples = new HashMap();
    listofmultiples.put(300, 0);
    listofmultiples.put(200, 0);
    listofmultiples.put(100, 0);
    listofmultiples.put(50, 0);
    int lastKey = 0;
    lastKey = (int) listofmultiples.keySet().toArray()[0];
    int i = listofmultiples.size();
    int valueforkey = 0;
    while (value > lastKey) {
        int key = (int) listofmultiples.keySet().toArray()[i - 1];
        if (value > key) {
            value %= key;
            valueforkey += 1;
            listofmultiples.put(key, valueforkey);
        }
        else{
            i-=1;
            valueforkey=0;
        }
    }
    System.out.println(listofmultiples.toString());
    System.out.println(value);
}
英文:

This code works for every number.

public static void main(String[] args) {
int value = 452;
Map&lt;Integer, Integer&gt; listofmultiples = new HashMap();
listofmultiples.put(300, 0);
listofmultiples.put(200, 0);
listofmultiples.put(100, 0);
listofmultiples.put(50, 0);
int lastKey = 0;
lastKey = (int) listofmultiples.keySet().toArray()[0];
int i = listofmultiples.size();
int valueforkey = 0;
while (value &gt; lastKey) {
int key = (int) listofmultiples.keySet().toArray()[i - 1];
if (value &gt; key) {
value %= key;
valueforkey += 1;
listofmultiples.put(key, valueforkey);
}
else{
i-=1;
valueforkey=0;
}
}
System.out.println(listofmultiples.toString());
System.out.println(value);
}

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  • 本文由 发表于 2020年9月14日 16:55:22
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