NumberFormatException，Java 中的 BigInteger
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NumberFormatException, BigInteger in java
提问by CoolEulerProject
Getting following run-time error
获得以下运行时错误
C:\jdk1.6.0_07\bin>java euler/BigConCheck
Exception in thread "main" java.lang.NumberFormatException: For input string: "z
"
at java.lang.NumberFormatException.forInputString(NumberFormatException.
java:48)
at java.lang.Integer.parseInt(Integer.java:447)
at java.math.BigInteger.<init>(BigInteger.java:314)
at java.math.BigInteger.<init>(BigInteger.java:447)
at euler.BigConCheck.conCheck(BigConCheck.java:25)
at euler.BigConCheck.main(BigConCheck.java:71)
My Code
我的代码
package euler;
import java.math.BigInteger;
class BigConCheck
{
public int[] conCheck(BigInteger big)
{
int i=0,q=0,w=0,e=0,r=0,t=0,mul=1;
int a[]= new int[1000];
int b[]= new int[7];
BigInteger rem[]= new BigInteger[4];
BigInteger num[]= new BigInteger[4];
for(i=0;i<4;i++)
num[i]=big; // intialised num[1 to 4][0] with big
String s="1",g="0";
for(i=0;i<999;i++)
s = s.concat(g);
BigInteger divi[]= new BigInteger[4];
for(i=0;i<5;i++)
{
divi[i]=new BigInteger(s);
int z = (int)Math.pow((double)10,(double)i);
BigInteger zz = new BigInteger("z"); // intialised div[1 to 4][0] with big
divi[i]=divi[i].divide(zz);
}
for(i=0;i<996;i++) // 5 consecative numbers.
{
for(int k=0;k<5;k++)
{
rem[k] = num[k].mod(divi[k]);
b[k]=rem[k].intValue();
mul= mul*b[k];
/*int z = (int)Math.pow((double)10,(double)(k+1));
String zz = "z";
BigInteger zzz = new BigInteger(zz);
num[k]=num[k].divide(zzz); */
}
a[i]=mul;
for(int p=0;p<5;p++)
{
BigInteger qq = new BigInteger("10");
num[p]=num[p].divide(qq);
}
}
return a;
}
public int bigestEleA(int u[])
{
int big=0;
for(int i=0;i<u.length;i++)
if(big<u[i])
big=u[i];
return big;
}
public static void main(String args[])
{
int con5[]= new int[1000];
int punCon;
BigInteger bigest = new BigInteger("7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450");
BigConCheck bcc = new BigConCheck();
con5=bcc.conCheck(bigest);
punCon=bcc.bigestEleA(con5);
System.out.println(punCon);
}
}
}
please point out whats goes wrong @ runtime and why
请指出@运行时出了什么问题以及为什么
thanks in advance...
提前致谢...
回答by nicholas.hauschild
This is the line causing you grief:
这是让你悲伤的线：
BigInteger zz = new BigInteger("z"); // intialised div[1 to 4][0] with big
While BigIntegerdoes work with String's, those String's must be parsable into numbers.
虽然BigInteger确实与String's一起工作，但那些String's 必须可解析为数字。
EDIT**Try this:
编辑**试试这个：
Integer z = (Integer)Math.pow((double)10,(double)i);
BigInteger zz = new BigInteger(z.toString());
回答by Bozho
new BigInteger("z");is not meaningful. You can only pass numbers in constructor.
new BigInteger("z");没有意义。您只能在构造函数中传递数字。
This is pretty obvious, so the next time you get an exception go the the exact line in your code shown in the exception stacktrace and you will most likely spot the problem.
这很明显，所以下次遇到异常时，请转到异常堆栈跟踪中显示的代码中的确切行，您很可能会发现问题。
回答by fmucar
BigInteger zz = new BigInteger("z");
you are passing non-numerical string thats the reason.
您正在传递非数字字符串，这就是原因。
EDIT:
编辑：
It takes string but it expects the string to be a numerical value. "z" does not have any numerical meaning.
它需要字符串，但它希望字符串是一个数值。“z”没有任何数字含义。
回答by Buhake Sindi
BigIntegerJavadoc states for BigInteger(String value)
BigIntegerJavadoc 声明为 BigInteger(String value)
Translates the decimal String representation of a BigInteger into a BigInteger.The String representation consists of an optional minus sign followed by a sequence of one or more decimal digits. The character-to-digit mapping is provided by Character.digit. The String may not contain any extraneous characters (whitespace, for example).
将 BigInteger 的十进制字符串表示形式转换为 BigInteger。字符串表示形式由一个可选的减号后跟一个或多个十进制数字的序列组成。字符到数字的映射由 Character.digit 提供。字符串不能包含任何多余的字符（例如，空格）。
So your code:
所以你的代码：
BigInteger zz = new BigInteger("z"); // intialised div[1 to 4][0] with big
is totally incorrect, but this is correct:
完全不正确，但这是正确的：
BigInteger zz = new BigInteger("5566");
EDIT: Based on your comment, this would be simpler by using the String.valueOf()method:
编辑：根据您的评论，使用以下String.valueOf()方法会更简单：
int z = (int)Math.pow((double)10,(double)i);
BigInteger zz = new BigInteger(String.valueOf(z));
回答by Pa?lo Ebermann
Could it be that you want this instead?
难道你想要这个吗？
int z = (int)Math.pow((double)10,(double)i);
BigInteger zz = new BigInteger(z);
Note the missing quotes here. (Of course, this will only work for i < 10.)
请注意此处缺少的引号。（当然，这只适用于i < 10。）
回答by Kris
A common mistake is writing new BigInteger("",num) instead of new BigInteger(""+num)
一个常见的错误是写 new BigInteger("",num) 而不是 new BigInteger(""+num)
