JAVA 大数据例题

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概述

用JAVA 实现大数的运算

还没有系统的学过java ,套用的模板

有几点需要注意的：

1、多组输入

while(cin.hasNext()){

}

2、判断输入的 a 是不是 0

在定义一个变量 c 首先将 c 赋值为 0

用 a.equals(c) 判断 a 是不是 0

3、清零

ans=new BigInteger（”0“）；

4、用 ans=ans.add(b); // 来计算 ans+ b 的值

eg 1:

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

Sample Output

370370367037037036703703703670

提交时package Bignum;不能交

package Bignum;
import java.util.*;
import java.math.*;
import java.math.BigInteger;
import java.util.Scanner;
 
public class Main {
	public static void main(String args[])
	{
		Scanner cin = new Scanner(System.in);
		BigInteger a;
		BigInteger ans;
		BigInteger c=BigInteger.valueOf(0);
		ans=new BigInteger("0");
		while(cin.hasNext())
		{
			a=cin.nextBigInteger();
			if(a.equals(c))
				break;
			else {
				ans=ans.add(a);
			}
		}
		System.out.println(ans);
		
	}
}

eg 2:

The problem is to multiply two integers X, Y . (0 ≤ X, Y < 10250)

Input

The input will consist of a set of pairs of lines. Each line in pair contains one multiplyer.

Output

For each input pair of lines the output line should consist one integer the product.

Sample Input

222222222222222222222222

Sample Output

144

444444444444444444444444

本题是实现大数的乘法乘法运算：

用到的是 ans=a.multiply(b); // 算 a*b

import java.util.*;
import java.math.*;
import java.math.BigInteger;
import java.util.Scanner;
 
public class Main {
	public static void main(String args[])
	{
		Scanner cin = new Scanner(System.in);
		BigInteger a,b,ans;
		while(cin.hasNext())
		{
			a=cin.nextBigInteger();
			b=cin.nextBigInteger();
			ans=a.multiply(b);
		    System.out.println(ans);
		}
		
	}
}

eg 3:

Sample Input
110 / 100
99 % 10
2147483647 / 2147483647
2147483646 % 2147483647
Sample Output
1
9
1
2147483646

import java.io.*;
import java.util.*;
import java.math.*;
import java.text.*;
public class Main {
 
	public static void main(String args[]) 
	{
		Scanner in = new Scanner(System.in);
        BigInteger a,b;char c;
        while(in.hasNext())
        {
		       a = in.nextBigInteger();
               c=in.next().charAt(0);
		       b = in.nextBigInteger();
		       if(c=='/')
                     System.out.println(a.divide(b));
               else
                     System.out.println(a.mod(b));
                       
         }
    }
}

eg 4:

次方

Sample Input
95.123
0.4321
5.1234
6.7592
98.999
1.0100
12
20
15
9
10
12
Sample Output
548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

import java.util.*;
import java.math.*;
public class Main
{
    public static void main(String args[])
    {
        Scanner cin=new Scanner(System.in);
        BigDecimal ans=new BigDecimal("1");
        BigDecimal a;
        while(cin.hasNext())
        {
        a=cin.nextBigDecimal();
        int n;
        n=cin.nextInt();
        ans=new BigDecimal("1");
        for(int i=1;i<=n;i++)
            ans=ans.multiply(a);
        String str;
        str=ans.stripTrailingZeros().toPlainString();//去除后导0
        if(str.charAt(0)=='0')
            System.out.println(str.substring(1));
        else
            System.out.println(str);
        }
    }
}

这个题没有用java

Write a program that reads an expression consisting of two non-negative integer and an operator.
Determine if either integer or the result of the expression is too large to be represented as a “normal”
signed integer (type integer if you are working Pascal, type int if you are working in C).
Input
An unspecified number of lines. Each line will contain an integer, one of the two operators ‘+’ or ‘*’,
and another integer.
Output
For each line of input, print the input followed by 0-3 lines containing as many of these three messages
as are appropriate: ‘first number too big’, ‘second number too big’, ‘result too big’.
Sample Input
300 + 3
9999999999999999999999 + 11
Sample Output
300 + 3
9999999999999999999999 + 11
first number too big
result too big

#include <cstdio>
#include <cstring>
#include <climits>
#include <cstdlib>
const int inf = INT_MAX;
int main()
{
    double x,y;
    char s1[1010],s2[10],s3[1010];
    while(~scanf("%s%s%s",s1,s2,s3))
    {
        printf("%s %c %sn",s1,s2[0],s3);
        x = atof(s1);
        y = atof(s3);
        if(s2[0] == '+')
        {
            if(x > inf)
                printf("first number too bign");
            if(y > inf)
                printf("second number too bign");
            if(x + y > inf)
                printf("result too bign");
        }
        if(s2[0] == '*')
        {
            if(x > inf)
                printf("first number too bign");
            if(y > inf)
                printf("second number too bign");
            if(x * y > inf)
                printf("result too bign");
        }
    }
    return 0;
}