HDU高精度总结（java大数类）
HDU1002 A + B Problem II
【题意】大数相加
【链接】http://acm.hdu.edu.cn/showproblem.php?pid=1002
Sample Input
2 1 2 112233445566778899 998877665544332211
Sample Output
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
代码：
import java.io.*;
import java.util.*;
import java.math.BigInteger;//声明BigInteger大数类
import java.lang.*;
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
int t,i=1;
t = cin.nextInt();
int tot = 0;
BigInteger a,b,c; //BigInteger类型
while (i<=t)
{
a=cin.nextBigInteger();
b=cin.nextBigInteger();
c=a.add(b);
System.out.println("Case "+i+":");
System.out.println(a+" + "+b+" = "+c);
if(iHDU1042 N!
【题意】大数阶乘
【链接】http://acm.hdu.edu.cn/showproblem.php?pid=1042
Sample Input
1 2 3
Sample Output
1 2 6
代码：
import java.io.*;
import java.util.*;
import java.math.BigInteger;//声明BigInteger大数类
import java.lang.*;
public class FF
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
while(cin.hasNext())
{
int n=cin.nextInt();
BigInteger ans=BigInteger.ONE;
for(int i=1; i<=n; ++i)
{
ans=ans.multiply(BigInteger.valueOf(i));
}
System.out.println(ans);
System.gc(); //调用垃圾回收机制
}
}
}【题意】多个大数相加
【链接】http://acm.hdu.edu.cn/showproblem.php?pid=1047
Sample Input
1 123456789012345678901234567890 123456789012345678901234567890 123456789012345678901234567890 0
Sample Output
370370367037037036703703703670
代码:
import java.io.*;
import java.util.*;
import java.math.BigInteger;//声明BigInteger大数类
import java.lang.*;
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
int n=cin.nextInt();
while(n-->0)
{
BigInteger a,b,c;
b=BigInteger.ZERO;
while(cin.hasNextBigInteger())
{
c=BigInteger.ZERO;
c=cin.nextBigInteger();
if(!c.equals(BigInteger.valueOf(0)))
b=b.add(c);
else
{
System.out.println(b);
if(n!=0)
System.out.println("");
break;
}
}
System.gc();
}
}
}
HDU 1715 大菲波数
【题意】RT
【链接】http://acm.hdu.edu.cn/showproblem.php?pid=1715
Sample Input
5 1 2 3 4 5
Sample Output
1 1 2 3 5
代码：
import java.io.*;
import java.util.*;
import java.math.BigInteger;//声明BigInteger大数类
import java.lang.*;
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
int n=cin.nextInt();
BigInteger fac[]= new BigInteger[1001];
fac[0]=BigInteger.ZERO;//初始赋值
fac[1]=BigInteger.ONE;
for(int i=2; i<=1000; ++i) fac[i]=fac[i-1].add(fac[i-2]);
while(n-->0)
{
int a;
a=cin.nextInt();
System.out.println(fac[a]);
}
//System.gc();
}
}
HDU 1063 Exponentiation
【题意】高精度幂
【链接】http://acm.hdu.edu.cn/showproblem.php?pid=1063
Sample Input
95.123 12 0.4321 20 5.1234 15 6.7592 9 98.999 10 1.0100 12
Sample Output
548815620517731830194541.899025343415715973535967221869852721 .00000005148554641076956121994511276767154838481760200726351203835429763013462401 43992025569.928573701266488041146654993318703707511666295476720493953024 29448126.764121021618164430206909037173276672 90429072743629540498.107596019456651774561044010001 1.126825030131969720661201
最简形式是去掉后面的 0，以及小于 1 的小数的小数点前的 0
实现高精度幂java方法：
（1）调用pow函数
（2）for循环
代码：
import java.io.*;
import java.util.*;
import java.math.BigDecimal;
import java.math.BigInteger;//声明BigInteger大数类
import java.lang.*;
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
while(cin.hasNext())
{
BigDecimal ans=cin.nextBigDecimal();
int n=cin.nextInt();
String res=ans.pow(n).stripTrailingZeros().toPlainString();//整数去掉小数点和后面的0 if(res.startsWith("0"))//去掉前导0
{
res=res.substring(1);
}
System.out.println(res);
/* BigDecimal a=BigDecimal.ONE;
int n=cin.nextInt();
for(int i=1; i<=n; ++i)
a=a.multiply(ans);
String res=a.stripTrailingZeros().toPlainString();
if(res.startsWith("0"))
{
res=res.substring(1);
}
System.out.println(res);
*/
}
}
}
HDU 1316 How Many Fibs?
【题意】区间fibonacci
【链接】http://acm.hdu.edu.cn/showproblem.php?pid=1316
Sample Input
10 100 1234567890 9876543210 0 0
Sample Output
5 4
代码：
import java.io.*;
import java.util.*;
import java.math.BigDecimal;
import java.math.BigInteger;//声明BigInteger大数类
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
BigInteger a,b;
int ans,i;
BigInteger fac[]=new BigInteger[1005];
BigInteger zero=BigInteger.ZERO;
fac[1]=BigInteger.valueOf(1);
fac[2]=BigInteger.valueOf(2);
for(i=3; i<1005; ++i) fac[i]=fac[i-1].add(fac[i-2]);
while(cin.hasNextBigInteger())
{
a=cin.nextBigInteger();
b=cin.nextBigInteger();
if(a.compareTo(zero)==0&&b.compareTo(zero)==0) break;
for(ans=0,i=1; i<1005; ++i)
{
if(a.compareTo(fac[i])<=0&&b.compareTo(fac[i])>=0) ans++;
if(b.compareTo(fac[i])<0) break;
}
System.out.println(ans);
}
}
}
【题意】高精度小数相加
【链接】http://acm.hdu.edu.cn/showproblem.php?pid=1753
Sample Input
1.1 2.9 1.1111111111 2.3444323343 1 1.1
Sample Output
4 3.4555434454 2.1
代码：
import java.io.*;
import java.util.*;
import java.math.BigDecimal;
import java.math.BigInteger;//声明BigInteger大数类
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
BigDecimal a,b,c;
while(cin.hasNextBigDecimal())
{
a=cin.nextBigDecimal();
b=cin.nextBigDecimal();
c=a.add(b);
String res=c.stripTrailingZeros().toPlainString();
System.out.println(res);
}
}
}