大数高精度之java处理

发布时间：2020-12-14 03:01:02 所属栏目：大数据来源：网络整理

导读：例题1 链接：点击打开链接原题： Octal Fractions Time Limit: ?1000MS ? Memory Limit: ?10000K Total Submissions: ?6648 ? Accepted: ?3627 Description Fractions in octal (base 8) notation can be expressed exactly in decimal notation. For examp

例题1

链接：点击打开链接

原题：

Octal Fractions

Time Limit:?1000MS	?	Memory Limit:?10000K
Total Submissions:?6648	?	Accepted:?3627

Description

Fractions in octal (base 8) notation can be expressed exactly in decimal notation. For example,0.75 in octal is 0.953125 (7/8 + 5/64) in decimal. All octal numbers of n digits to the right of the octal point can be expressed in no more than 3n decimal digits to the right of the decimal point.?

Write a program to convert octal numerals between 0 and 1,inclusive,into equivalent decimal numerals.

Input

The input to your program will consist of octal numbers,one per line,to be converted. Each input number has the form 0.d1d2d3 ... dk,where the di are octal digits (0..7). There is no limit on k.

Output

Your output will consist of a sequence of lines of the form?

0.d1d2d3 ... dk [8] = 0.D1D2D3 ... Dm [10]

where the left side is the input (in octal),and the right hand side the decimal (base 10) equivalent. There must be no trailing zeros,i.e. Dm is not equal to 0.

Sample Input

0.75
0.0001
0.01234567

Sample Output

0.75 [8] = 0.953125 [10]
0.0001 [8] = 0.000244140625 [10]
0.01234567 [8] = 0.020408093929290771484375 [10]

Source

Southern African 2001

java.source:

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

public class da {
public static void main(String args[])
{
BigDecimal a,t,c,ans,tmp;
String s;
Scanner cin=new Scanner(new BufferedInputStream(System.in));
t=BigDecimal.valueOf(8);
while(cin.hasNextLine())
{
s=cin.nextLine();
tmp=t;
ans=BigDecimal.ZERO;
for(int i=2;i<s.length();i++)
{
c=BigDecimal.valueOf(s.charAt(i)-'0');
ans=ans.add(c.divide(tmp));
tmp=tmp.multiply(t);
}
System.out.println(s+" [8] = "+ans+" [10]");
}

}

}

例题2

链接：点击打开链接

原题：

Water Treatment Plants

Time Limit:?1000MS	?	Memory Limit:?10000K
Total Submissions:?798	?	Accepted:?478

Description

River polution control is a major challenge that authorities face in order to ensure future clean water supply. Sewage treatment plants are used to clean-up the dirty water comming from cities before being discharged into the river.?

As part of a coordinated plan,a pipeline is setup in order to connect cities to the sewage treatment plants distributed along the river. It is more efficient to have treatment plants running at maximum capacity and less-used ones switched off for a period. So,each city has its own treatment plant by the river and also a pipe to its neighbouring city upstream and a pipe to the next city downstream along the riverside. At each city's treatment plant there are three choices:?

either process any water it may receive from one neighbouring city,together with its own dirty water,discharging the cleaned-up water into the river;?
or send its own dirty water,plus any from its downstream neighbour,along to the upstream neighbouring city's treatment plant (provided that city is not already using the pipe to send it's dirty water downstream);?
or send its own dirty water,plus any from the upstream neighbour,to the downstream neighbouring city's plant,if the pipe is not being used.?

The choices above ensure that:?

every city must have its water treated somewhere and?
at least one city must discharge the cleaned water into the river.?
Let's represent a city discharging water into the river as "V" (a downwards flow),passing water onto its neighbours as ">" (to the next city on its right) or else "<" (to the left). When we have several cities along the river bank,we assign a symbol to each (V,< or >) and list the cities symbols in order. For example,two cities,A and B,can?

each treat their own sewage and each discharges clean water into the river. So A's action is denoted V as is B's and we write "VV" ;?
or else city A can send its sewage along the pipe (to the right) to B for treatment and discharge,denoted ">V";?
or else city B can send its sewage to (the left to) A,which treats it with its own dirty water and discharges (V) the cleaned water into the river. So A discharges (V) and B passes water to the left (<),and we denote this situation as "V<".?
We could not have "><" since this means A sends its water to B and B sends its own to A,so both are using the same pipe and this is not allowed. Similarly "<<" is not possible since A's "<" means it sends its water to a non-existent city on its left.?

So we have just 3 possible set-ups that fit the conditions:?

         A    B       A > B       A < B 

         V    V           V       V             

  RIVER~ ~ ~ ~ ~     ~ ~ ~ ~ ~   ~ ~ ~ ~ ~RIVER

          "VV"        ">V"         "V<"

If we now consider three cities,we can determine 8 possible set-ups.?
Your task is to produce a program that given the number of cities NC (or treatment plants) in the river bank,determines the number of possible set-ups,NS,that can be made according to the rules define above.?

You need to be careful with your design as the number of cities can be as large as 100.?

Input

The input consists of a sequence of values,where each value represents the number of cities.

Output

Your output should be a sequence of values,where each value represents the number of possible set-ups for the corresponding number of cities read in the same input line.

Sample Input

2
3
20

Sample Output

3
8
102334155

Source

Southwestern Europe 2002

题意思路转自 http://blog.sina.com.cn/s/blog_6635898a0100ovby.html

题意：有n个城市，它们由一个污水处理系统连接着，每个城市可以选择
1、将左边城市过来的污水和右边城市过来的污水连同本身的污水排到河里 ?>V<
2、将左边来的污水连同自己的污水排到右边??>>
3、将右边来的污水连同自己的污水排到左边??<<

问总共有多少种处理情况，即不同又符合实际的><V组合。

思路：DP+高精度。DP部分，易得最右边城市的状态只可能用3种：>V,V,<。故分三种状态讨论，设dp[i][0]为第i个城市的状态为：> V ，dp[i][1]为：V ，dp[i][2]为：<。由实际流动的可能性可以得到状态转移方程：

dp[i][0] = dp[i-1][0] + dp[i-1][1]；

dp[i][1] = dp[i-1][0] + dp[i-1][1] + dp[i-1][2];

dp[i][2] = dp[i-1][0] + dp[i-1][1] + dp[i-1][2];

然后可以整理为：dp[i] = 3 * dp[i-1] -dp[i-2]。然后用java的BigInteger预处理就OK了。

My java.source：

import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void main(String args[])
{
Scanner cin=new Scanner(new BufferedInputStream(System.in));
BigInteger dp[]=new BigInteger[111];
dp[1]=BigInteger.valueOf(1);
dp[2]=BigInteger.valueOf(3);
for(int i=3;i<=100;i++)
{
dp[i]=BigInteger.valueOf(3).multiply(dp[i-1]).subtract(dp[i-2]);
}
while(cin.hasNext())
{
int n;
n=cin.nextInt();
System.out.println(dp[n]);
}

}

}

例题3

链接：点击打开链接

原题：

NUMBER BASE CONVERSION

Time Limit:?1000MS	?	Memory Limit:?10000K
Total Submissions:?4322	?	Accepted:?1949

Description

Write a program to convert numbers in one base to numbers in a second base. There are 62 different digits:?
{ 0-9,A-Z,a-z }?
HINT: If you make a sequence of base conversions using the output of one conversion as the input to the next,when you get back to the original base,you should get the original number.?

Input

The first line of input contains a single positive integer. This is the number of lines that follow. Each of the following lines will have a (decimal) input base followed by a (decimal) output base followed by a number expressed in the input base. Both the input base and the output base will be in the range from 2 to 62. That is (in decimal) A = 10,B = 11,...,Z = 35,a = 36,b = 37,z = 61 (0-9 have their usual meanings).?

Output

The output of the program should consist of three lines of output for each base conversion performed. The first line should be the input base in decimal followed by a space then the input number (as given expressed in the input base). The second output line should be the output base followed by a space then the input number (as expressed in the output base). The third output line is blank.?

Sample Input

8
62 2 abcdefghiz
10 16 1234567890123456789012345678901234567890
16 35 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
35 23 333YMHOUE8JPLT7OX6K9FYCQ8A
23 49 946B9AA02MI37E3D3MMJ4G7BL2F05
49 61 1VbDkSIMJL3JjRgAdlUfcaWj
61 5 dl9MDSWqwHjDnToKcsWE1S
5 10 42104444441001414401221302402201233340311104212022133030

Sample Output

62 abcdefghiz
2 11011100000100010222220010010110022222001001100011010010001

10 1234567890123456789012345678901234567890
16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2

16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
35 333YMHOUE8JPLT7OX6K9FYCQ8A

35 333YMHOUE8JPLT7OX6K9FYCQ8A
23 946B9AA02MI37E3D3MMJ4G7BL2F05

23 946B9AA02MI37E3D3MMJ4G7BL2F05
49 1VbDkSIMJL3JjRgAdlUfcaWj

49 1VbDkSIMJL3JjRgAdlUfcaWj
61 dl9MDSWqwHjDnToKcsWE1S

61 dl9MDSWqwHjDnToKcsWE1S
5 42104444441001414401221302402201233340311104212022133030

5 42104444441001414401221302402201233340311104212022133030
10 1234567890123456789012345678901234567890

Source

Greater New York 2002

题意：大数的进制转换。将a进制的正整数num转换为b进制的正整数。

思路：高精度，java，要注意为num = 0的情况。

题意思路注释转自:http://blog.sina.com.cn/s/blog_6635898a0100ovel.html

My java.source:

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

public class Main {
public static void main(String[] args)
{
Scanner cin=new Scanner(System.in);
int cas,i,w;
BigInteger sum,b1,b2;
char c;
cas=cin.nextInt();
while((cas--)!=0)
{
b1=cin.nextBigInteger();//??读入大数。
b2=cin.nextBigInteger();
String s=cin.next(); ? ? ? ? ?// ?读入字符串，无空格，而cin.nextline()有。
sum=BigInteger.valueOf(0);
t=BigInteger.valueOf(1);
for(i=s.length()-1;i>=0;i--)?先将s转换为10进制。
{
c=s.charAt(i);
if(c>='0'&&c<='9')
w=c-'0';
else if(c>='A'&&c<='Z')
w=c-'A'+10;
else
w=c-'a'+36;
sum=sum.add(BigInteger.valueOf(w).multiply(t));
t=t.multiply(b1);
}
int top=0;
int stack[]=new int[200];
while(sum.compareTo(BigInteger.valueOf(0))!=0)?转化为b2进制的数，存在stack[]中。
{
stack[++top]=sum.mod(b2).intValue();
sum=sum.divide(b2);
}
System.out.print(b1+" "+s+"n"+b2+" ");
if(top==0)
System.out.print(0);?要注意为0的情况。
while(top!=0)
{
w=stack[top--];
if(w<10)
c=(char)(w+'0');
else if(w<36)
c=(char)(w-10+'A');
else
c=(char)(w-36+'a');
System.out.print(c);
}
System.out.print("nn");
}
}
}

例题4

链接：点击打开链接

原题：

Heritage

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?7129	?	Accepted:?2645

Description

Your rich uncle died recently,and the heritage needs to be divided among your relatives and the church (your uncle insisted in his will that the church must get something). There are N relatives (N <= 18) that were mentioned in the will. They are sorted in descending order according to their importance (the first one is the most important). Since you are the computer scientist in the family,your relatives asked you to help them. They need help,because there are some blanks in the will left to be filled. Here is how the will looks:?

Relative #1 will get 1 / ... of the whole heritage,?
Relative #2 will get 1 / ... of the whole heritage,?
---------------------- ...?
Relative #n will get 1 / ... of the whole heritage.?

The logical desire of the relatives is to fill the blanks in such way that the uncle's will is preserved (i.e the fractions are non-ascending and the church gets something) and the amount of heritage left for the church is minimized.

Input

The only line of input contains the single integer N (1 <= N <= 18).

Output

Output the numbers that the blanks need to be filled (on separate lines),so that the heritage left for the church is minimized.

Sample Input

Sample Output

2
3

Source

ural 1108

题意：将一份遗产分成n份，1/X1,1/X2,1/X3,1/Xn。使得X1 >= X2 >= X3 >= ... >= Xn，都为正整数。最后剩下的分给教堂，让其分得的财产最少，但不能没有。

思路：贪心+高精度。首先由贪心的思想可以知道，如果剩下1/6的遗产给最后一人和教堂，易得最后一人的得到的遗产为1/7。由这个思想可以列出 1/(Xn-1) = 1 - 1/X1 - 1/X2 -...- 1/X(n-1)。然后这个式子和X(n+1)的式子相消，化简，可得X(n+1) = Xn * (Xn - 1) + 1。然后就用java的大数A掉。

题意思路注释转自: http://blog.sina.com.cn/s/blog_6635898a0100ovom.html

My java source:

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

public class Main{
public static void main(String args[])
{
int i,n,cas;
BigInteger ans;
Scanner cin=new Scanner(System.in);
while(cin.hasNext())
{
ans=BigInteger.valueOf(2);
n=cin.nextInt();
for(i=1;i<=n;i++)
{
System.out.println(ans);
ans=ans.multiply(ans.subtract(BigInteger.valueOf(1))).add(BigInteger.valueOf(1));
}
}
}
}

例题5

链接：点击打开链接

原题：

Integer Inquiry

Time Limit:?1000MS	?	Memory Limit:?10000K
Total Submissions:?30091	?	Accepted:?11718

Description

One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.?
``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were here to see these results.'' (Chip moved to a new apartment,once one became available on the third floor of the Lemon Sky apartments on Third Street.)?

Input

The input will consist of at most 100 lines of text,each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length,and will only contain digits (no VeryLongInteger will be negative).?

The final input line will contain a single zero on a line by itself.?

Output

Your program should output the sum of the VeryLongIntegers given in the input.

Sample Input

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

Sample Output

370370367037037036703703703670

Source

East Central North America 1996

My java.source:

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

public class Main{
public static void main(String args[])
{
BigInteger num,ans;
Scanner cin=new Scanner(System.in);
ans=BigInteger.valueOf(0);
while(cin.hasNext())
{
num=cin.nextBigInteger();
if(num.compareTo(BigInteger.valueOf(0))==0)
break;
ans=ans.add(num);
}
System.out.println(ans);
}
}

例题6

链接：点击打开链接

原题：

Just the Facts

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?8747	?	Accepted:?4645

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

Source

South Central USA 1997

java.source:

import java.io.*; import java.util.*; import java.math.*; import java.text.*; public class Main { public static void main(String args[]) { BigInteger n,nn; String s; int i; Scanner cin=new Scanner(System.in); while(cin.hasNext()) { n=cin.nextBigInteger(); nn=BigInteger.valueOf(1); for(i=2;i<=n.intValue();i++) { nn=nn.multiply(BigInteger.valueOf(i)); } s=nn.toString(); for(i=s.length()-1;i>=0;i--) { if(s.charAt(i)!='0') { break; } } System.out.printf("%5d -> %cn",s.charAt(i)); } } }

（编辑：李大同）

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