hdu1042 N!(大数求阶乘)
发布时间:2020-12-14 04:30:43 所属栏目:大数据 来源:网络整理
导读:N! Time Limit: 10000/5000 MS (Java/Others)????Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 94583????Accepted Submission(s): 28107 Problem Description Given an integer N(0 ≤ N ≤ 10000),your task is to calculate N! ? ?
N!Time Limit: 10000/5000 MS (Java/Others)????Memory Limit: 262144/262144 K (Java/Others)
Problem Description
Given an integer N(0 ≤ N ≤ 10000),your task is to calculate N!
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Input
One N in one line,process to the end of file.
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Output
For each N,output N! in one line.
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Sample Input
1
2
3
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Sample Output
1
2
6
数据范围比较大,要用到大数。 c++ 1 #include<bits/stdc++.h> 2 using namespace std; 3 string Multiply(string s,int x) 4 { 5 reverse(s.begin(),s.end()); 6 int cmp=0; 7 for(int i=0; i<s.size(); i++) 8 { 9 cmp=(s[i]-‘0‘)*x+cmp; 10 s[i]=(cmp%10+‘0‘); 11 cmp/=10; 12 } 13 while(cmp) 14 { 15 s+=(cmp%10+‘0‘); 16 cmp/=10; 17 } 18 reverse(s.begin(),s.end()); 19 return s; 20 } 21 int main() 22 { 23 int n; 24 while(~scanf("%d",&n)) 25 { 26 string sum="1"; 27 for(int i=1; i<=n; i++) 28 { 29 sum=Multiply(sum,i); 30 } 31 cout<<sum<<endl; 32 } 33 return 0; 34 } Java 1 import java.math.BigInteger; 2 import java.util.Scanner; 3 public class Main { 4 public static void main(String[] args) { 5 Scanner cin = new Scanner(System.in ); 6 while(cin.hasNext()) { 7 int n=cin.nextInt(); 8 BigInteger sum=BigInteger.valueOf(1); 9 for(int i=1;i<=n;i++) 10 { 11 BigInteger temp=BigInteger.valueOf(i); 12 sum=sum.multiply(temp); 13 } 14 System.out.println(sum); 15 16 } 17 } 18 } PS:这题好像还是Java比较快。。。 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |