poj 1604 Just the Facts

最新推荐文章于 2017-05-29 20:49:07 发布

原创最新推荐文章于 2017-05-29 20:49:07 发布 · 748 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法 #poj #编程 #c++

OJ编程专栏收录该内容

81 篇文章

订阅专栏

本文介绍了一种计算特定范围内任意非负整数阶乘的最后一个非零数字的方法。通过预先计算并存储每个整数阶乘的最后一个非零数字，实现快速查询输出。文章提供了一个完整的C++代码示例，利用了模运算来处理乘法结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Just the Facts

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8781		Accepted: 4659

Description

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

For this problem, you are to write a program that can compute the last non-zero digit of any factorial for (0 <= N <= 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

题意：求阶乘n！的最后一个不为0的数字；

解题思路：直接模拟打表，注意每次做乘法得到的值需要模除100000；最后输出格式需要注意；

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
#define MAX 10000
int main(){
	int n,a[MAX+5];
	int mul=1;
	memset(a,0,sizeof(a));
	for (int i=1;i<MAX;i++){
		mul*=i;
		while (mul%10==0)
			mul/=10;
		mul%=100000;
		a[i]=mul%10;
	}
	while (cin>>n){
		printf("%5d -> %d\n",n,a[n]);
	}
	return 0;
}