uva 568 Just the Facts(f分解因子）

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本文介绍了一种计算特定范围内整数阶乘后最后一个非零数字的方法，并提供了两种实现方案，一种通过直接计算并去除尾部零的方式，另一种则采用因子分解技巧，确保了计算精度。

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Just the Facts

The expression N!, read as ``N factorial," denotes the product of the first N positive integers, where N is nonnegative. So, for example,

N	N!
`0`	`1`
`1`	`1`
`2`	`2`
`3`	`6`
`4`	`24`
`5`	`120`
`10`	`3628800`

For this problem, you are to write a program that can compute the last non-zero digit of any factorial for ( $0 \le N \le 10000$ ). For example, if your program is asked to compute the last nonzero digit of 5!, your program should produce ``2" because 5! = 120, and 2 is the last nonzero digit of 120.

Input

Input to the program is a series of nonnegative integers not exceeding 10000, each on its own line with no other letters, digits or spaces. For each integer N, you should read the value and compute the last nonzero digit of N!.

Output

For each integer input, the program should print exactly one line of output. Each line of output should contain the value N, right-justified in columns 1 through 5 with leading blanks, not leading zeroes. Columns 6 - 9 must contain `` -> " (space hyphen greater space). Column 10 must contain the single last non-zero digit of N!.

Sample Input

Sample Output

题目大意：给出n(n < 10000),求n!后面有多少个0。

解题思路：每乘一个数的时候就要进行判断，去除后面的0，因为n<10000，所以每次可以模掉10000.（10000刚好，1000会有误差）

#include<stdio.h>

int main(){
	int n;
	while (scanf("%d", &n) != EOF){
		int sum = 1;
		for (int i = 1; i <= n; i++){
			sum = sum * i;
			while (sum % 10 == 0){
				sum /= 10;
			}
			sum %= 100000;
		}
		printf("%5d -> %d\n", n, sum % 10);
	}
	return 0;
}

贴个牛叉的代码，用到分解因子。

#include<stdio.h>
int main()
{
   int n,i,num;
   while(scanf("%d",&n)!=EOF)
   {
      num=1;
      for(i=1;i<=n;i++)
      {
        if(i%5==0)
        {
           int x=i;
           while(x%10==0)
              x/=10;
           while(x%5==0)
           {
              x/=5;
              num/=2;
           }
           num*=x;
         }
         else
            num*=i;
         num%=100000;//一定要至少保留后五位非零位(五位以上要注意int型溢出，因此取五位最佳），因为后五位非零位的进位（极限情况3125，即5^5）会影响到最后的非零位
    }
   printf("%5d -> %d\n",n,num%10);
 }
return 0;
}