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当使用 Java 使用整数计算 100 (100!) 的阶乘时，我得到 0

发布于 2024-10-22 06:30:31 字数 478 浏览 6 评论 0原文

这样做时：

int x = 100;
int result = 1;
for (int i = 1; i < (x + 1); i++) {
    result = (result * i);
}
System.out.println(result);

这显然是因为结果对于整数来说太大了，但我习惯于为溢出获取大负数，而不是 0。

提前感谢！

当我切换到这个时：

int x = 100;
int result = 1;

for (int i = 1; i < (x + 1); i++) {
    result = (result * i);
    System.out.println(result);
}

我得到this。

原文

When doing this:

int x = 100;
int result = 1;
for (int i = 1; i < (x + 1); i++) {
    result = (result * i);
}
System.out.println(result);

This is clearly because the result is too big for an integer, but I am used to get big negative numbers for the overflow, and not 0.

Thanks in advance!

When I switch to this:

int x = 100;
int result = 1;

for (int i = 1; i < (x + 1); i++) {
    result = (result * i);
    System.out.println(result);
}

I get this.

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评论（10）

蓝天 2024-10-29 06:30:31

1到100之间有50个偶数。这意味着阶乘是 2 的倍数至少 50 次，换句话说，作为二进制数，最后 50 位将是 0。（实际上更重要，因为偶数第二个偶数是 2*2 的倍数等

public static void main(String... args) {
    BigInteger fact = fact(100);
    System.out.println("fact(100) = " + fact);
    System.out.println("fact(100).longValue() = " + fact.longValue());
    System.out.println("fact(100).intValue() = " + fact.intValue());
    int powerOfTwoCount = 0;
    BigInteger two = BigInteger.valueOf(2);
    while (fact.compareTo(BigInteger.ZERO) > 0 && fact.mod(two).equals(BigInteger.ZERO)) {
        powerOfTwoCount++;
        fact = fact.divide(two);
    }
    System.out.println("fact(100) powers of two = " + powerOfTwoCount);
}

private static BigInteger fact(long n) {
    BigInteger result = BigInteger.ONE;
    for (long i = 2; i <= n; i++)
        result = result.multiply(BigInteger.valueOf(i));
    return result;
}

）

fact(100) = 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
fact(100).longValue() = 0
fact(100).intValue() = 0
fact(100) powers of two = 97

这意味着对于fact(100) 的最低位来说，97 位整数将为0。

事实上，对于fact(n) 来说，2 的幂数非常接近n。事实上（10000）有 9995 个 2 的幂。这是因为它大约是 1/2 的 n 次幂之和，总计接近 n。即每第二个数字都是偶数 n/2，每 4 个数字都有一个额外的 2 次幂 (+n/4)，每一个第 8 个数字有一个额外的幂 (+n/8) 等等，接近 n 作为总和。

There are 50 even numbers between 1 and 100 inclusive. This means that the factorial is a multiple of 2 at least 50 times, in other words as a binary number the last 50 bits will be 0. (Actually it is more as even second even number is a multiple of 2*2 etc)

public static void main(String... args) {
    BigInteger fact = fact(100);
    System.out.println("fact(100) = " + fact);
    System.out.println("fact(100).longValue() = " + fact.longValue());
    System.out.println("fact(100).intValue() = " + fact.intValue());
    int powerOfTwoCount = 0;
    BigInteger two = BigInteger.valueOf(2);
    while (fact.compareTo(BigInteger.ZERO) > 0 && fact.mod(two).equals(BigInteger.ZERO)) {
        powerOfTwoCount++;
        fact = fact.divide(two);
    }
    System.out.println("fact(100) powers of two = " + powerOfTwoCount);
}

private static BigInteger fact(long n) {
    BigInteger result = BigInteger.ONE;
    for (long i = 2; i <= n; i++)
        result = result.multiply(BigInteger.valueOf(i));
    return result;
}

prints

fact(100) = 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
fact(100).longValue() = 0
fact(100).intValue() = 0
fact(100) powers of two = 97

This means a 97-bit integer would be 0 for the lowest bits of fact(100)

In fact, the number of powers of two is very close to n for fact(n). For fact(10000) there are 9995 powers of two. This is because its is approximately the sum of n times powers of 1/2 giving a total close to n. i.e. every second number is even n/2 and every 4th has an additional power of 2 (+n/4) and every 8th has an additional power (+n/8) etc approaches n as a sum.

回复 原文
云淡风轻 2024-10-29 06:30:31

大负数是溢出到特定范围的值； factorial(100) 末尾有超过 32 个二进制零，因此将其转换为整数会产生零。

Big negative numbers are values that overflowed into certain ranges; factorial(100) has more than 32 binary zeros on the end, so converting it to an integer produces zero.

回复 原文
晚雾 2024-10-29 06:30:31

为了了解原因，我们可以观察阶乘的素因式分解。

fac( 1) = 1             = 2^0
fac( 2) = 2             = 2^1
fac( 3) = 2 * 3         = 2^1 * 3
fac( 4) = 2 * 2 * 2 * 3 = 2^3 * 3
fac( 5) =  ...          = 2^3 * 3 * 5
fac( 6) = ...           = 2^4 * 3^2 * 5
fac( 7) = ...           = 2^4 * ...
fac( 8) = ...           = 2^7 * ...
fac( 9) = ...           = 2^7 * ...
fac(10) = ...           = 2^8 * ...
fac(11) = ...           = 2^8 * ...
...
fac(29) = ...           = 2^25 * ...
fac(30) = ...           = 2^26 * ...
fac(31) = ...           = 2^26 * ...
fac(32) = ...           = 2^31 * ...
fac(33) = ...           = 2^31 * ...
fac(34) = ...           = 2^32 * ...  <===
fac(35) = ...           = 2^32 * ...
fac(36) = ...           = 2^34 * ...
...
fac(95) = ...           = 2^88 * ...
fac(96) = ...           = 2^93 * ...
fac(97) = ...           = 2^93 * ...
fac(98) = ...           = 2^94 * ...
fac(99) = ...           = 2^94 * ...
fac(100)= ...           = 2^96 * ...

2 的指数是基数 2 视图中尾随零的数量，因为所有其他因子都是奇数，因此在最后一个二进制数字中贡献一个 1产品。

类似的方案也适用于其他素数，因此我们可以轻松计算 fac(100) 的因式分解：

fac(100) = 2^96 * 3^48 * 5^24 * 7^16 * 11^9 * 13^7 * 17^5 * 19^5 * 23^4 *
           29^3 * 31^2 * 37^2 * 41^2 * 43^2 * 47^2 *
           53 * 59 * 61 * 67 * 71 * 73 * 79 * 83 * 89 * 97

因此，如果我们的计算机以 3 为基数存储数字，并且有 48 个 trit-numbers ，fac(100) 将为 0（也如 fac(99)，但 fac(98) 不会:-)

To have a look at the cause, we could observe the prime factorization of the factorial.

fac( 1) = 1             = 2^0
fac( 2) = 2             = 2^1
fac( 3) = 2 * 3         = 2^1 * 3
fac( 4) = 2 * 2 * 2 * 3 = 2^3 * 3
fac( 5) =  ...          = 2^3 * 3 * 5
fac( 6) = ...           = 2^4 * 3^2 * 5
fac( 7) = ...           = 2^4 * ...
fac( 8) = ...           = 2^7 * ...
fac( 9) = ...           = 2^7 * ...
fac(10) = ...           = 2^8 * ...
fac(11) = ...           = 2^8 * ...
...
fac(29) = ...           = 2^25 * ...
fac(30) = ...           = 2^26 * ...
fac(31) = ...           = 2^26 * ...
fac(32) = ...           = 2^31 * ...
fac(33) = ...           = 2^31 * ...
fac(34) = ...           = 2^32 * ...  <===
fac(35) = ...           = 2^32 * ...
fac(36) = ...           = 2^34 * ...
...
fac(95) = ...           = 2^88 * ...
fac(96) = ...           = 2^93 * ...
fac(97) = ...           = 2^93 * ...
fac(98) = ...           = 2^94 * ...
fac(99) = ...           = 2^94 * ...
fac(100)= ...           = 2^96 * ...

The exponent for the 2 is the number of trailing zeros in the base-2 view, as all other factors are odd, and thus contribute a 1 in the last binary digit to the product.

A similar scheme works for other prime numbers, too, so we can easily calculate the factorization of fac(100):

fac(100) = 2^96 * 3^48 * 5^24 * 7^16 * 11^9 * 13^7 * 17^5 * 19^5 * 23^4 *
           29^3 * 31^2 * 37^2 * 41^2 * 43^2 * 47^2 *
           53 * 59 * 61 * 67 * 71 * 73 * 79 * 83 * 89 * 97

So, if our computer stored the numbers in base 3, and had 48-trit-numbers, fac(100) would be 0 (as fac(99), too, but fac(98) would not :-)

回复 原文
颜漓半夏 2024-10-29 06:30:31

好问题 - 答案是：
33 的阶乘（由于负值）为 -2147483648，即 0x80000000，如果采用 64 位，则为 0xFFFFFFFF80000000。乘以 34（下一个成员）将得到一个 long 值 0xFFFFFFE600000000，当转换为 int 时将得到 0x00000000。

显然从那时起你将保持 0。

Nice problem - answer is:
Factorial of 33 (due to negative values) is -2147483648 which is 0x80000000, or 0xFFFFFFFF80000000 if taking 64bits. Multiplying by 34 (the next member) will give a long value of 0xFFFFFFE600000000, which when casting to int will give you 0x00000000.

Obviously from that point onwards you will remain with 0.

回复 原文
静若繁花 2024-10-29 06:30:31

使用递归和 BigIntegers 的简单解决方案：

    public static BigInteger factorial(int num){
    if (num<=1)
        return BigInteger.ONE;
    else
        return factorial(num-1).multiply(BigInteger.valueOf(num));
    }

输出：

93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

Simple solution using recursion and BigIntegers:

    public static BigInteger factorial(int num){
    if (num<=1)
        return BigInteger.ONE;
    else
        return factorial(num-1).multiply(BigInteger.valueOf(num));
    }

Output:

93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
回复 原文
乖乖 2024-10-29 06:30:31

（在此处找到，稍微修改以适应问题）

public static void main(String[] args) {

    BigInteger fact = BigInteger.valueOf(1);
    for (int i = 1; i <= 100; i++)
        fact = fact.multiply(BigInteger.valueOf(i));
    System.out.println(fact);
}

(Found here, adapted slightly to fit question)

public static void main(String[] args) {

    BigInteger fact = BigInteger.valueOf(1);
    for (int i = 1; i <= 100; i++)
        fact = fact.multiply(BigInteger.valueOf(i));
    System.out.println(fact);
}
回复 原文
仙女 2024-10-29 06:30:31

Java 中的 BigInteger 类。 BigInteger 类用于数学运算，涉及超出所有可用原始数据类型限制的非常大的整数计算。

要计算非常大的数字，我们可以使用BigInteger

就像，如果我们想计算 45 的阶乘，
答案=119622220865480194561963161495657715064383733760000000000

 static void extraLongFactorials(int n) {
       BigInteger fact = BigInteger.ONE;
        for(int i=2; i<=n; i++){
            fact = fact.multiply(BigInteger.valueOf(i));
        }
        System.out.println(fact);
    }

BigInteger的主要方法是BigInteger.ONE，BigInteger.ZERO，BigInteger.TEN，BigInteger.ValueOf()

BigInteger Class in Java. BigInteger class is used for mathematical operation which involves very big integer calculations that are outside the limit of all available primitive data types.

To calculate very large number, we can use BigInteger

Like, If we want to calculate the factorial of 45,
answer = 119622220865480194561963161495657715064383733760000000000

 static void extraLongFactorials(int n) {
       BigInteger fact = BigInteger.ONE;
        for(int i=2; i<=n; i++){
            fact = fact.multiply(BigInteger.valueOf(i));
        }
        System.out.println(fact);
    }

Main methods of BigInteger is BigInteger.ONE, BigInteger.ZERO, BigInteger.TEN, BigInteger.ValueOf()

回复 原文
潇烟暮雨 2024-10-29 06:30:31
import java.util.*;
import java.math.*;
public class BigInteger_Factorial {
    public static void main(String args []){
        Scanner s = new Scanner(System.in);

        BigInteger x,i,fac = new BigInteger("1");
        x = s.nextBigInteger();

        for(i=new BigInteger("1"); i.compareTo(x)<=0; i=i.add(BigInteger.ONE)){
            fac = fac.multiply((i));
        }
        System.out.println(fac);
    }
}

输出 100 作为输入：

93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

输出图像：

Output result

import java.util.*;
import java.math.*;
public class BigInteger_Factorial {
    public static void main(String args []){
        Scanner s = new Scanner(System.in);

        BigInteger x,i,fac = new BigInteger("1");
        x = s.nextBigInteger();

        for(i=new BigInteger("1"); i.compareTo(x)<=0; i=i.add(BigInteger.ONE)){
            fac = fac.multiply((i));
        }
        System.out.println(fac);
    }
}

Output of 100 as input:

93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

Output image:

Output result

回复 原文
纸短情长 2024-10-29 06:30:31
package test2;

import java.math.BigInteger;
import java.util.Scanner;

public class Factorial extends Big {

    public static void main(String args []){ 
    int x,fact=1,i ;
    Scanner sc = new Scanner(System.in);
    
    
    System.out.println("press any dight and 0 to exit");
    while (sc.nextInt()!=0)
    {
    System.out.println("Enter the values ");
    x=sc.nextInt();
    if(x<26)

    {
    for( i=1;i<=x;i++)
    {   fact = fact*i;  }
    
    System.out.println("Factorial of "+x + "is "+ fact );
    
    fact=1;
    }
    else 
    {
        System.out.println("In else big....");
    BigInteger k=fact(x);
    
    System.out.println("The factorial of "+x+"is "+k);
    System.out.println("RESULT LENGTH\n"+k.toString().length());
    }
    System.out.println("press any dight and 0 to exit");
    }
    System.out.println("thanks....");
    }
    
        
    
    
}
//----------------------------------------------------//

package test2;

import java.math.BigInteger;

public class Big {

    public static void main(String... args) {
        BigInteger fact = fact(100);
        System.out.println("fact(100) = " + fact);
        System.out.println("fact(100).longValue() = " + fact.longValue());
        System.out.println("fact(100).intValue() = " + fact.intValue());
        int powerOfTwoCount = 0;
        BigInteger two = BigInteger.valueOf(2);
        while (fact.compareTo(BigInteger.ZERO) > 0 && fact.mod(two).equals(BigInteger.ZERO)) {
            powerOfTwoCount++;
            fact = fact.divide(two);
        }
        System.out.println("fact(100) powers of two = " + powerOfTwoCount);
    }

    public static BigInteger fact(long n) {
        BigInteger result = BigInteger.ONE;
        for (long i = 2; i <= n; i++)
            result = result.multiply(BigInteger.valueOf(i));
        return result;
    }
    
}   

package test2;

import java.math.BigInteger;
import java.util.Scanner;

public class Factorial extends Big {

    public static void main(String args []){ 
    int x,fact=1,i ;
    Scanner sc = new Scanner(System.in);
    
    
    System.out.println("press any dight and 0 to exit");
    while (sc.nextInt()!=0)
    {
    System.out.println("Enter the values ");
    x=sc.nextInt();
    if(x<26)

    {
    for( i=1;i<=x;i++)
    {   fact = fact*i;  }
    
    System.out.println("Factorial of "+x + "is "+ fact );
    
    fact=1;
    }
    else 
    {
        System.out.println("In else big....");
    BigInteger k=fact(x);
    
    System.out.println("The factorial of "+x+"is "+k);
    System.out.println("RESULT LENGTH\n"+k.toString().length());
    }
    System.out.println("press any dight and 0 to exit");
    }
    System.out.println("thanks....");
    }
    
        
    
    
}
//----------------------------------------------------//

package test2;

import java.math.BigInteger;

public class Big {

    public static void main(String... args) {
        BigInteger fact = fact(100);
        System.out.println("fact(100) = " + fact);
        System.out.println("fact(100).longValue() = " + fact.longValue());
        System.out.println("fact(100).intValue() = " + fact.intValue());
        int powerOfTwoCount = 0;
        BigInteger two = BigInteger.valueOf(2);
        while (fact.compareTo(BigInteger.ZERO) > 0 && fact.mod(two).equals(BigInteger.ZERO)) {
            powerOfTwoCount++;
            fact = fact.divide(two);
        }
        System.out.println("fact(100) powers of two = " + powerOfTwoCount);
    }

    public static BigInteger fact(long n) {
        BigInteger result = BigInteger.ONE;
        for (long i = 2; i <= n; i++)
            result = result.multiply(BigInteger.valueOf(i));
        return result;
    }
    
}
回复 原文
寻梦旅人 2024-10-29 06:30:31

这肯定是溢出，你可以尝试双精度，64位长整数可能太小

it's an overflow for sure, you can try double, 64 bit long integer are probably too small

回复 原文
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