题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1042[代码]
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2021-08-04 16:15:16
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#include<stdio.h>
#include<iostream>
using namespace std;
int a[8000];
int main()
{int n;int i;while(scanf("%d",&n)!=EOF)//碰到有EOF的时候一定要用scanf。不要用cin>>n&&n这样的的，不让这样的错误你非
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2017-05-19 20:45:00
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#include<iostream>
#include<stdio.h>
#include<string>
#include<iomanip>
#include<algorithm>
using namespace std;
#include "time.h"
const int MAX_GROUPS = 20000;//最多2
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2021-07-28 16:45:11
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题目链接：http://acm.
hdu.edu.cn/showproblem.php?pid=
1042
#include<iostream> #include<stdio.h> #include<string> #include<iomanip> #include<algorithm>
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2009-09-14 16:23:00
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问题链接：HDU1042 n!。问题描述：参见上文。问题分析：看似简单的问题，背后实际上有许多道理。阶乘值随着n的增大，增大的速度相当的快，是一个大整数。似乎这个题可以使用大整数类进行计算，然而类计算相对复杂，时间上溢出的可能性比较大。一般的整数计算，在计算机中是一种...
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2016-05-29 09:51:00
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N!Time Limit: 10000/5000 MS (Java/Others) Me
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2022-08-08 18:04:19
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题目大意: 求 n。(可能会超过整数范围，这里用数组模拟n！的值) http://acm.hdu.edu.cn/showproblem.php?pid=1042 AC代码: /** *数组模拟n！,循环太多可能超时， */ #include<iostream> #include<cstdio> #i
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2021-08-06 18:07:16
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我的代码，由于时间的问题，以后找机会优化了！
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2021-07-28 16:49:01
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Problem Description Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N! Input One N in one line, process to the end of file. Output For ea
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2021-07-14 16:00:54
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题解：高精度乘法压位。#include #include const int mod=10000;int l,n,a[1000000];int main(){ while(~scanf("%d",&n)){ memset(a,0,sizeof a); a[1]=1;...
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2014-08-29 15:24:00
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N!Time Limit: 10000/5000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 40204Accepted Submission(s): 11162 Problem DescriptionGiven an integer N(0 ≤ N ≤ 10000), your task is to calculate N!InputOne N in one line, process to the end of file.OutputFor each N, output N!
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2013-03-30 10:08:00
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N!http://acm.hdu.edu.cn/showproblem.php?pid=1042Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 37977 Accepted Submission(s): 10555 Problem DescriptionGiven an integer N(0 ≤ N ≤ 10000), your task is to calculate N!InputOne N in one line, process
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2012-12-26 16:55:00
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N!
Time Limit: 10000/5000 MS (J
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2023-07-17 15:52:52
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//123*20 相当于 100*20 + 20*20+3 //常规方法N>=13就溢出 #include#incl
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2012-07-30 14:58:00
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N! Problem Description Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N! Input One N in one line, process to the end of file. Output For eac
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2013-11-01 19:32:00
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2011-12-23 07:49:22地址：http://acm.hdu.edu.cn/showproblem.php?pid=1042题意：算n!。最高10000。1w的时候n!有不超过4w位(Log10(10000^10000) = 40000)。代码：# include <stdio.h># define MOD 10000int num[40000] ;void mul(int n){ int i, cc = 0 ; for (i = 1 ; i <= num[0] ; i++) { num[i] = num[i]*n + cc ; cc ...
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2012-01-06 23:29:00
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N!Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33314 Accepted Submission(s): 9234Problem Description Given an integer N(0 ≤ N ≤ 10000), your task is to calculate NInput One N in one line, process to the end of file.Output For each N, output N!
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2012-07-23 18:28:00
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N!
Problem Description
Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!
Input
One N in one line, process to the end of file.
Sample Input
1
2
3
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2023-08-15 17:21:41
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http://acm.hdu.edu.cn/showproblem.php?pid=1042集训第一天比赛上有这个题 以前做过一个求150的 效率比较低 用在这个上面就超时 然后在刘汝佳书上看到一种方法 勉强可以过 3400+msView Code 1 #include<stdio.h> 2 #include<string.h> 3 int x[40000]; 4 int main() 5 { 6 int i, n,g,s,d; 7 while(scanf("%d", &n)!=EOF) 8 { 9 int max = 40000;10 ..
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2012-07-18 16:05:00
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题目大意：输入n，输出n!解题思路：数据可能很大。用java的大数代码如下：package com.njupt.bigInteger;import java.math.BigInteger;import java.util.Scanner;public class HDU_1042 { public static void main(String[] ar
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2013-08-11 20:09:38
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