UVa 568 Just the Facts (数论&打表&不打表)

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本文介绍了一种计算任意非负整数阶乘后尾部第一个非零数字的方法。通过两种不同的算法实现：一种是预计算并存储结果的打表法；另一种是通过质因数分解计算的直接算法。这两种方法都能高效地解决UVa在线评测系统中的568号问题。

568 - Just the Facts

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=100&page=show_problem&problem=509

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

方法1：打表。根据数据范围，计算时保留最后5位就行。

方法2：计算N!的素因子分解中5的幂的个数count，然后能被2整除的数要除以2直到达到count

先放上打表的算法：

/*0.016s*/

#include<cstdio>

int a[10001] = {1};/// 0!=1

int main()
{
	int i, x;
	for (i = 1; i <= 10000; i++)
	{
		a[i] = a[i - 1] * i;
		while (a[i] % 10 == 0) a[i] /= 10;
		a[i] %= 100000;///根据数据范围，保留最后5位就行~
	}
	while (~scanf("%d", &x))
		printf("%5d -> %d\n", x, a[x] % 10);
	return 0;
}

然后是不打表的算法：

/*0.016s*/

#include<cstdio>

int p[10000];

int main()
{
	int n;
	while (~scanf("%d", &n))
	{
		int i, sum = 1, count = 0, tem = 0;
		if (n == 0)
		{
			printf("    0 -> 1\n");
			continue;
		}
		for (i = 0; i < n; i++) p[i] = i + 1;
		for (i = 0; i < n; i++)
			while (true)
			{
				if (p[i] % 5 == 0)
				{
					p[i] /= 5;
					count++;
				}
				else break;
			}
		for (i = 0; i < n; i++)
		{
			while (true)
			{
				if (p[i] % 2 == 0)
				{
					p[i] /= 2;
					tem++;
				}
				if (p[i] % 2 || tem == count)
					break;
			}
			if (tem == count)
				break;
		}
		for (i = 0; i < n; i++)
		{
			sum *= p[i];
			sum %= 10;
		}
		printf("%5d -> %d\n", n, sum);
	}
	return 0;
}