UVa 568 - Just the Facts

UVa568-JusttheFacts代码实现与解析

最新推荐文章于 2025-11-29 10:44:03 发布

最新推荐文章于 2025-11-29 10:44:03 发布 · 87 阅读

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本文详细介绍了UVa568-JusttheFacts问题的解题思路、代码实现及关键参考链接。通过计算任意非负整数的阶乘最后非零数字，展示了如何避免溢出并高效求解。

UVa 568 - Just the Facts

1 题目
2 思路
3 代码
4 参考

1题目

===================

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

Miguel A. Revilla
1998-03-10

===================

2思路

题目大意是计算N!最后一个非0位，为了避免溢出，代码中使用了模运算。为了取得非0位，一旦计算结果是10的倍数，就将末位的0通过除法消除。

3代码

/*
 * Problem: UVa 568 - Just the Facts
 * Lang: ANSI C
 * Time: 0.019
 * Author: minix
 */

#include <stdio.h>

#define N 100000000000

int main() {
  long long sum;
  int i, n;

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

  return 0;
}