Java 阶乘函数为21产生错误的结果!及以上
因此,将创建以下输出:Java 阶乘函数为21产生错误的结果!及以上,java,long-integer,factorial,integer-overflow,Java,Long Integer,Factorial,Integer Overflow,因此,将创建以下输出: public static long factorial(int num1) { if (num1 == 0) return 1; else return Math.abs(num1 * factorial(num1 - 1)); } 结果是21分!是错误的(它应该是51090942171709440000),结果变成完全混乱的22!及以上。有人能解释为什么吗?对于第21个值及以上的值,值变得不稳定,因为真实值对于长的来说
public static long factorial(int num1)
{
if (num1 == 0)
return 1;
else
return Math.abs(num1 * factorial(num1 - 1));
}
结果是21分!是错误的(它应该是51090942171709440000),结果变成完全混乱的22!及以上。有人能解释为什么吗?对于第21个值及以上的值,值变得不稳定,因为真实值对于
长的
来说太大了。如果您需要更大的数字,请使用。在long
容量范围内,这一切似乎都可以。你为什么要使用Math.abs()?它并没有明显的不稳定。看见你20岁的时候还不错。在此之后,我将开始查看long的最大值(9223372036854775807)。一个long
在Java中可以接受的最大值是9223372036854775807,它在20之后溢出代码>
您应该使用BigInteger来计算阶乘
例如:
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5,040
8! = 40,320
9! = 362,880
10! = 3,628,800
11! = 39,916,800
12! = 479,001,600
13! = 6,227,020,800
14! = 87,178,291,200
15! = 1,307,674,368,000
16! = 20,922,789,888,000
17! = 355,687,428,096,000
18! = 6,402,373,705,728,000
19! = 121,645,100,408,832,000
20! = 2,432,902,008,176,640,000
21! = 4,249,290,049,419,214,848
22! = 1,250,660,718,674,968,576
23! = 8,128,291,617,894,825,984
24! = 7,835,185,981,329,244,160
25! = 7,034,535,277,573,963,776
biginger n=biginger.ONE;
对于(int i=1;i和long,您可以表示介于-9223372036854775808到+9223372036854775807()因此,一旦阶乘通过该范围,您就开始出现错误。小纠正:22!
的结果是第一个错误的结果。它表明22!<21!
这显然是不正确的。我猜是因为OP注意到在24之后的数字会变成负值。
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5,040
8! = 40,320
9! = 362,880
10! = 3,628,800
11! = 39,916,800
12! = 479,001,600
13! = 6,227,020,800
14! = 87,178,291,200
15! = 1,307,674,368,000
16! = 20,922,789,888,000
17! = 355,687,428,096,000
18! = 6,402,373,705,728,000
19! = 121,645,100,408,832,000
20! = 2,432,902,008,176,640,000
21! = 4,249,290,049,419,214,848
22! = 1,250,660,718,674,968,576
23! = 8,128,291,617,894,825,984
24! = 7,835,185,981,329,244,160
25! = 7,034,535,277,573,963,776
BigInteger n = BigInteger.ONE;
for (int i=1; i<=20; i++) {
n = n.multiply(BigInteger.valueOf(i));
System.out.println(i + "! = " + n);
}