英文:
Last 3 non zero digits from factorial
问题
可以有人帮我用Java编写示例代码，演示如何从大阶乘中找出最后3个非零数字？
例如：12! ：479001600 = 16
10! ：3628800 = 288
英文:
Can someone help me with the sample code in java how to find last 3 non zero digits from big factorial?
Eg 12! :- 479001600 = 16
10! :- 3628800 =288
答案1
得分: 1
以下函数计算阶乘：
private static BigInteger factorial(int n) {
return IntStream.rangeClosed(1, n)
.mapToObj(BigInteger::valueOf)
.collect(Collectors.reducing(BigInteger.ONE, BigInteger::multiply));
}
而这个函数计算最后三位非零数字：
private static BigInteger last3NonzeroDigits(BigInteger n) {
while (n.mod(BigInteger.TEN).equals(BigInteger.ZERO)) {
n = n.divide(BigInteger.TEN);
}
return n.mod(BigInteger.valueOf(1000));
}
- 删除尾随零：当最后一位是0（可被10整除，即 i % 10 = 0），则除以10
- 从结果数字中，提取最多的最后三位数字（i % 1000）
测试：
for (int i = 1; i <= 15; i++) {
BigInteger f = factorial(i);
System.out.println(i + "! = " + f + " -> " + last3NonzeroDigits(f));
}
输出：
1! = 1 -> 1
2! = 2 -> 2
3! = 6 -> 6
4! = 24 -> 24
5! = 120 -> 12
6! = 720 -> 72
7! = 5040 -> 504
8! = 40320 -> 32
9! = 362880 -> 288
10! = 3628800 -> 288
11! = 39916800 -> 168
12! = 479001600 -> 16
13! = 6227020800 -> 208
14! = 87178291200 -> 912
15! = 1307674368000 -> 368
英文:
The following function calculates the factorial:
private static BigInteger factorial(int n) {
return IntStream.rangeClosed(1, n)
.mapToObj(BigInteger::valueOf)
.collect(Collectors.reducing(BigInteger.ONE, BigInteger::multiply));
}
And this function calculates the last 3 non-zero digits:
private static BigInteger last3NonzeroDigits(BigInteger n) {
while (n.mod(BigInteger.TEN).equals(BigInteger.ZERO)) {
n = n.divide(BigInteger.TEN);
}
return n.mod(BigInteger.valueOf(1000));
}
- delete trailing zeroes: while the last digit is 0 (dividable by 10, hence i % 10 = 0), divide by 10
- from the resulting number, extract the (at most) last 3 digits (i % 1000)
Test:
for (int i = 1; i <= 15; i++) {
BigInteger f = factorial(i);
System.out.println(i+"! = "+f + " -> " + last3NonzeroDigits(f));
}
Output:
1! = 1 -> 1
2! = 2 -> 2
3! = 6 -> 6
4! = 24 -> 24
5! = 120 -> 12
6! = 720 -> 72
7! = 5040 -> 504
8! = 40320 -> 32
9! = 362880 -> 288
10! = 3628800 -> 288
11! = 39916800 -> 168
12! = 479001600 -> 16
13! = 6227020800 -> 208
14! = 87178291200 -> 912
15! = 1307674368000 -> 368
答案2
得分: 0
以下是翻译好的内容：
你可以将数字映射到字符串，循环遍历各个数字，找到最后一个位置，其中可以找到最大的三个非零数字，最后可以返回该索引，并将最后的数字打印出来作为你的结果。我已经编写了一点代码，并编写了方法 findLastIndex 来获取索引：
fun findLastIndex(num: String): Int {
var zero = true
var counter = 0
val reversedNum = num.reversed()
for(i in 0 until num.length){
if(!reversedNum[i].equals('0')){
counter++
zero = false;
}
if((reversedNum[i].equals('0') || counter >= 3) && !zero){
return num.length - i - 1
}
}
return 0
}
现在你可以调用这个方法并打印出最后的非零数字：
val num = 479001600
val numString = num.toString()
val index = findLastIndex(numString)
println(numString.substring(index).replace("0", ""))
你可以在 kotlin playground 中测试它。在 Java 中进行映射应该很容易实现。关于 reverse 方法，你可以参考下面的 文章。或者你也可以颠倒 for 循环。希望对你有所帮助。
英文:
You can map the number to a string, loop over the digits, find the last index where maximum three non-zero digits can be found and at the end you can return the index and print the last digits as your result. I have coded a little bit and write the method findLastIndex to get the index:
fun findLastIndex(num: String): Int {
var zero = true
var counter = 0
val reversedNum = num.reversed()
for(i in 0 until num.length){
if(!reversedNum[i].equals('0')){
counter++
zero = false;
}
if((reversedNum[i].equals('0') || counter >= 3) && !zero){
return num.length - i - 1
}
}
return 0
}
Now you can call the method and print the last non-zero digits:
val num = 479001600
val numString = num.toString()
val index = findLastIndex(numString)
println(numString.substring(index).replace("0", ""))
You can test it in the kotlin playground. Mapping in Java should be easy to do. For the reverse method you can have a look at the following article. Or you can inverse the for loop. Hope it helps.
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