（部分转载，部分原创）java大数类（2）

NYOJ 773  开方数

http://acm.nyist.net/JudgeOnline/problem.php?pid=773

import java.util.Scanner;

public class Main{
    public static void main(String[] args){
        int n;
        double p;
        Scanner cin = new Scanner(System.in);
        while(true){
            n = cin.nextInt();
            p = cin.nextDouble();
            if(n == 0 && p == 0) break;
            System.out.printf("%.0f
", Math.pow(p, 1.0/n));
        }
    }
}

 

POJ 1001 Exponentiation

http://poj.org/problem?id=1001

import java.io.*;
import java.math.*;
import java.util.*;
import java.text.*;

public class Main 
{
    public static void main(String[] args) 
    {
        Scanner cin = new Scanner(System.in);
        while(cin.hasNext())
        {
       BigDecimal R = cin.nextBigDecimal();
       int n = cin.nextInt();
         String ans = R.pow(n).stripTrailingZeros().toPlainString();
         if(ans.startsWith("0"))
             ans = ans.substring(1);
         System.out.println(ans);
       }
    }
}

 

HDU 1042  N!

http://acm.hdu.edu.cn/showproblem.php?pid=1042

java版：

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import java.util.Scanner;
import java.math.BigInteger;
public class Main{
    public static void main(String[] args){
        Scanner input = new Scanner(System.in);
        int A, i;
        while(input.hasNext())
        {    
            A = input.nextInt();
            BigInteger B = BigInteger.ONE;
            for(i=1; i<=A; i++)
            {
                B = B.multiply(BigInteger.valueOf(i));
            }
            System.out.println(B);
        }
    }
}

C++版：

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#include<iostream>
#include<string.h>
#include<stdio.h>
#include<ctype.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<math.h>
#include<vector>
#include<deque>
#include<list>
#define INF 0x7fffffff
#define inf 0x3f3f3f3f
#define maxn 40000
using namespace std;
int a[maxn];
int main()
{
    int n;
    int s;
    while(scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        a[0]=1;
        for(int i=2; i<=n; i++)
        {
            int c=0;
            for(int j=0; j<maxn; j++)
            {
                s=a[j]*i+c;
                a[j]=s%10;
                c=s/10;
            }
        }
        int p;
        for(int i=maxn; i>=0; i--)
            if(a[i])
            {
                p=i;
                break;
            }
        for(int i=p; i>=0; i--)
            printf("%d",a[i]);
        printf("
");
    }
    return 0;
}

 

 

FZOJ 1602  Best results

http://acm.fzu.edu.cn/problem.php?pid=1602

import java.io.*;
import java.math.*;
import java.util.*;
import java.text.*;

public class Main 
{
    public static void main(String[] args) 
    {
        Scanner cin = new Scanner(System.in);
        while(cin.hasNext())
        {
          int t = cin.nextInt();
          BigDecimal ans = BigDecimal.ONE;
          while(t-- > 0)
            {
           String str = cin.next();
           int l = str.length();
           str = str.substring(0, l - 1);
           int tep = Integer.parseInt(str);
          BigDecimal Tep = BigDecimal.valueOf(tep);
           Tep = Tep.divide(BigDecimal.valueOf(100));
           ans = ans.multiply(Tep);
            }
          System.out.println(ans.stripTrailingZeros().toPlainString());
        }
    }
}

 

 

ZOJ  3549   Little Keng

一开始超时了，百撕不得骑姐，后来意识到可能JAVA计算大数和小数的时候，复杂度还是差别挺大的，虽然理论上都是O(1)的操作，但是当数的规模增长到一定级数，运算效率的差别就显露出来了。

不超时的关键就是快速幂的时候加上取模的操作。

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3549

import java.math.*;
import java.util.*;

public class Main {
    static BigInteger power(BigInteger p,BigInteger n,BigInteger m)            //快速幂
    {
        BigInteger sq=BigInteger.ONE;
        while(n.compareTo(BigInteger.ZERO)>0)                    //while(n>0)
        {
            if(n.mod(BigInteger.valueOf(2)).compareTo(BigInteger.ONE)==0)    //if(n%2==1)
                sq=sq.multiply(p).mod(m);                //    sq=(sq*p)%m;
            p=p.multiply(p).mod(m);                        //p=(p*p)%m;
            n=n.divide(BigInteger.valueOf(2));                //n=n/2;
        }
        return sq.mod(m);
    }
    static boolean judge(int m,int n,int k)
    {
        BigInteger mod=BigInteger.ONE,ans=BigInteger.ZERO;
        int i;
        for(i=1;i<=k;i++)                            //10^k
            mod=mod.multiply(BigInteger.valueOf(10));
        for(i=1;i<=m;i++)                            //1^n+2^n+3^n+...+m^n
        {
            BigInteger a,b;
            a=BigInteger.valueOf(i);
            b=BigInteger.valueOf(n);
            ans=(ans.add(power(a,b,mod))).mod(mod);
        }
        if(ans.mod(mod).compareTo(BigInteger.ZERO)==0) return true;
        else return false;
    }
    public static void main(String args[])
    {
        Scanner cin=new Scanner(System.in);
        int i,m,n;
        while(cin.hasNext())
        {
            m=cin.nextInt();
            n=cin.nextInt();
            for(i=1;;i++)
            {
                if(judge(m,n,i)) continue;
                else break;
            }
            System.out.println(i-1);
        }
    }
}

这是正解。。。

 

import java.io.*;
import java.math.*;
import java.util.*;
import java.text.*;

class Main 
{
   /* static BigInteger Pow(BigInteger a, int b)
    {
        BigInteger ans = BigInteger.ONE;
        while(b > 0)
        {
            if(b % 2 == 1)
                ans = ans.multiply(a);
            ans = ans.multiply(ans);
            b >>= 1;
            //b /= 2; 
        }
        return ans;
    } */


     static BigInteger power(BigInteger p,BigInteger n)         //快速幂  
    {  
        BigInteger sq=BigInteger.ONE;  
        while(n.compareTo(BigInteger.ZERO)>0)                    //while(n>0)  
        {  
            if(n.mod(BigInteger.valueOf(2)).compareTo(BigInteger.ONE)==0)   //if(n%2==1)  
                sq=sq.multiply(p);               //  sq=(sq*p)%m;  
            p=p.multiply(p);                     //p=(p*p)%m;  
            n=n.divide(BigInteger.valueOf(2));              //n=n/2;  
        }  
        return sq;  
    } 

    public static void main(String[] args) 
    {
        Scanner cin = new Scanner (System.in);
        while(cin.hasNext())
        {
            int m = cin.nextInt();
            int n = cin.nextInt();

            //System.out.println(Pow(BigInteger.TEN, 1));

            BigInteger ans = BigInteger.ZERO;
            for(int i = 1; i <= m; ++i)
            {
                BigInteger I = BigInteger.valueOf(i);
                BigInteger N = BigInteger.valueOf(n);
                ans = ans.add(power(I, N));
              //ans = ans.add(Pow(I, n));
            }
            int cnt = 0;
            while(ans.mod(BigInteger.TEN).compareTo(BigInteger.ZERO) == 0)
            {
                ans = ans.divide(BigInteger.TEN);
                ++cnt;
            }
                
          System.out.println(cnt);   
        }
        //System.out.println("Hello World!");
    }
}

这是TLE的代码。。。