Java大数类处理的简单程序
2013-04-24 17:53
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大数阶乘
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=28
代码如下:
[java] view plaincopyprint?
1. import java.io.*;
2. import java.math.BigInteger;
3. import java.util.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner cin = new Scanner(System.in);
10. int n = cin.nextInt();
11. BigInteger ans = BigInteger.ONE;
12. for(int i = 1; i <= n; ++i)
13. ans = ans.multiply(BigInteger.valueOf(i));
14. System.out.println(ans);
15. }
16. }
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
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);
}
}
棋盘覆盖
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=45
代码如下:
[java] view plaincopyprint?
1. import java.math.BigInteger;
2. import java.util.*;
3. import java.io.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner in = new Scanner(System.in);
10. int test = in.nextInt();
11. while(test-- > 0)
12. {
13. int n;
14. n = in.nextInt();
15. BigInteger a = new BigInteger("4");
16. for(int i = 1; i < n; ++i)
17. a = a.multiply(BigInteger.valueOf(4));
18. System.out.println(a.subtract(BigInteger.valueOf(1)).divide(BigInteger.valueOf(3)));
19. }
20. }
21. }
import java.math.BigInteger;
import java.util.*;
import java.io.*;
public class Main
{
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
int test = in.nextInt();
while(test-- > 0)
{
int n;
n = in.nextInt();
BigInteger a = new BigInteger("4");
for(int i = 1; i < n; ++i)
a = a.multiply(BigInteger.valueOf(4));
System.out.println(a.subtract(BigInteger.valueOf(1)).divide(BigInteger.valueOf(3)));
}
}
}
比较大小
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=73
代码如下:
[java] view plaincopyprint?
1. import java.io.*;
2. import java.math.BigInteger;
3. import java.util.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner cin = new Scanner(System.in);
10. while(cin.hasNext())
11. {
12. BigInteger a = cin.nextBigInteger();
13. BigInteger b = cin.nextBigInteger();
14. if(a.equals(BigInteger.ZERO) && b.equals(BigInteger.ZERO))
15. break;
16. int flag = a.compareTo(b);
17. if(flag == -1)
18. System.out.println("a<b");
19. else if(flag == 0)
20. System.out.println("a==b");
21. else
22. System.out.println("a>b");
23. }
24. }
25. }
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
while(cin.hasNext())
{
BigInteger a = cin.nextBigInteger();
BigInteger b = cin.nextBigInteger();
if(a.equals(BigInteger.ZERO) && b.equals(BigInteger.ZERO))
break;
int flag = a.compareTo(b);
if(flag == -1)
System.out.println("a<b");
else if(flag == 0)
System.out.println("a==b");
else
System.out.println("a>b");
}
}
}
大数加法
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=103
代码如下:
[java] view plaincopyprint?
1. import java.math.BigInteger;
2. import java.util.*;
3. import java.io.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner in = new Scanner(System.in);
10. int n = in.nextInt();
11. for(int i = 1; i <= n; ++i)
12. {
13. BigInteger a = in.nextBigInteger();
14. BigInteger b = in.nextBigInteger();
15. BigInteger ans = a.add(b);
16. System.out.println("Case " + i + ":");
17. System.out.println(a + " + " + b + " = " +ans);
18. }
19. }
20. }
import java.math.BigInteger;
import java.util.*;
import java.io.*;
public class Main
{
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
int n = in.nextInt();
for(int i = 1; i <= n; ++i)
{
BigInteger a = in.nextBigInteger();
BigInteger b = in.nextBigInteger();
BigInteger ans = a.add(b);
System.out.println("Case " + i + ":");
System.out.println(a + " + " + b + " = " +ans);
}
}
}
递推求值
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=114
代码如下:
[java] view plaincopyprint?
1. import java.io.*;
2. import java.math.BigInteger;
3. import java.util.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner cin = new Scanner(System.in);
10. BigInteger a[] = new BigInteger[100];
11. while(cin.hasNext())
12. {
13. for(int i = 0; i <= 2; ++i)
14. a[i] = cin.nextBigInteger();
15. for(int i = 3; i <= 99; ++i)
16. a[i] = a[i - 1].add(a[i - 2]).add(a[i - 3]);
17. System.out.println(a[99]);
18. }
19. }
20. }
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
BigInteger a[] = new BigInteger[100];
while(cin.hasNext())
{
for(int i = 0; i <= 2; ++i)
a[i] = cin.nextBigInteger();
for(int i = 3; i <= 99; ++i)
a[i] = a[i - 1].add(a[i - 2]).add(a[i - 3]);
System.out.println(a[99]);
}
}
}
高精度幂
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=155
代码如下:
[java] view plaincopyprint?
1. import java.io.*;
2. import java.math.BigDecimal;
3. import java.util.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner cin = new Scanner(System.in);
10. while(cin.hasNext())
11. {
12. BigDecimal ans = cin.nextBigDecimal();
13. int n = cin.nextInt();
14. String res = ans.pow(n).stripTrailingZeros().toPlainString(); //整数去掉小数点和后面的0
15. if(res.startsWith("0")) //去掉前导0
16. {
17. res = res.substring(1);
18. }
19. System.out.println(res);
20. }
21. }
22. }
计算100!+101!+……+200!
import java.math.*;
public class No0305
{
public static void main(String args[])
{
BigInteger
first100=new BigInteger("100"),//初始化各个大整数
last201=new BigInteger("201"),
one=new BigInteger("1"),
power=new BigInteger("1"),
result=new BigInteger("0");
for (BigInteger i=one;i.compareTo(last201)<0;i=i.add(one))//循环累加从1到201
{
power=power.multiply(i);//求i!
if(i.compareTo(first100)>=0)//从100!开始加
result=result.add(power);
}
System.out.println(result.toString());
}
}
补充:
public static void main(String[] args) {
//int的最大数值是2的31次方减1,如果在最大值上继续增加则会从最小值开始,循环下去。
//BigInteger没有最大值,但计算并不是很方便需要使用里面定义的各种方法才可以。
Integer i = new Integer(2147483647);
BigInteger bi = new BigInteger("2147483647");
System.out.println(i+1);
System.out.println(bi.add(BigInteger.ONE));
}
java中数值型数据的 无穷大、无穷小 的写法
无穷大:
Integer.MAX_VALUE;
Long.MAX_VALUE;
Double.MAX_VALUE
无穷小:
Integer.MIN_VALUE;
Long.MIN_VALUE;
Double.MIN_VALUE;
参考网址:http://www.cnblogs.com/Lyush/category/291376.html
http://blog.csdn.net/niushuai666/article/details/6972991
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=28
代码如下:
[java] view plaincopyprint?
1. import java.io.*;
2. import java.math.BigInteger;
3. import java.util.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner cin = new Scanner(System.in);
10. int n = cin.nextInt();
11. BigInteger ans = BigInteger.ONE;
12. for(int i = 1; i <= n; ++i)
13. ans = ans.multiply(BigInteger.valueOf(i));
14. System.out.println(ans);
15. }
16. }
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
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);
}
}
棋盘覆盖
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=45
代码如下:
[java] view plaincopyprint?
1. import java.math.BigInteger;
2. import java.util.*;
3. import java.io.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner in = new Scanner(System.in);
10. int test = in.nextInt();
11. while(test-- > 0)
12. {
13. int n;
14. n = in.nextInt();
15. BigInteger a = new BigInteger("4");
16. for(int i = 1; i < n; ++i)
17. a = a.multiply(BigInteger.valueOf(4));
18. System.out.println(a.subtract(BigInteger.valueOf(1)).divide(BigInteger.valueOf(3)));
19. }
20. }
21. }
import java.math.BigInteger;
import java.util.*;
import java.io.*;
public class Main
{
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
int test = in.nextInt();
while(test-- > 0)
{
int n;
n = in.nextInt();
BigInteger a = new BigInteger("4");
for(int i = 1; i < n; ++i)
a = a.multiply(BigInteger.valueOf(4));
System.out.println(a.subtract(BigInteger.valueOf(1)).divide(BigInteger.valueOf(3)));
}
}
}
比较大小
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=73
代码如下:
[java] view plaincopyprint?
1. import java.io.*;
2. import java.math.BigInteger;
3. import java.util.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner cin = new Scanner(System.in);
10. while(cin.hasNext())
11. {
12. BigInteger a = cin.nextBigInteger();
13. BigInteger b = cin.nextBigInteger();
14. if(a.equals(BigInteger.ZERO) && b.equals(BigInteger.ZERO))
15. break;
16. int flag = a.compareTo(b);
17. if(flag == -1)
18. System.out.println("a<b");
19. else if(flag == 0)
20. System.out.println("a==b");
21. else
22. System.out.println("a>b");
23. }
24. }
25. }
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
while(cin.hasNext())
{
BigInteger a = cin.nextBigInteger();
BigInteger b = cin.nextBigInteger();
if(a.equals(BigInteger.ZERO) && b.equals(BigInteger.ZERO))
break;
int flag = a.compareTo(b);
if(flag == -1)
System.out.println("a<b");
else if(flag == 0)
System.out.println("a==b");
else
System.out.println("a>b");
}
}
}
大数加法
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=103
代码如下:
[java] view plaincopyprint?
1. import java.math.BigInteger;
2. import java.util.*;
3. import java.io.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner in = new Scanner(System.in);
10. int n = in.nextInt();
11. for(int i = 1; i <= n; ++i)
12. {
13. BigInteger a = in.nextBigInteger();
14. BigInteger b = in.nextBigInteger();
15. BigInteger ans = a.add(b);
16. System.out.println("Case " + i + ":");
17. System.out.println(a + " + " + b + " = " +ans);
18. }
19. }
20. }
import java.math.BigInteger;
import java.util.*;
import java.io.*;
public class Main
{
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
int n = in.nextInt();
for(int i = 1; i <= n; ++i)
{
BigInteger a = in.nextBigInteger();
BigInteger b = in.nextBigInteger();
BigInteger ans = a.add(b);
System.out.println("Case " + i + ":");
System.out.println(a + " + " + b + " = " +ans);
}
}
}
递推求值
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=114
代码如下:
[java] view plaincopyprint?
1. import java.io.*;
2. import java.math.BigInteger;
3. import java.util.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner cin = new Scanner(System.in);
10. BigInteger a[] = new BigInteger[100];
11. while(cin.hasNext())
12. {
13. for(int i = 0; i <= 2; ++i)
14. a[i] = cin.nextBigInteger();
15. for(int i = 3; i <= 99; ++i)
16. a[i] = a[i - 1].add(a[i - 2]).add(a[i - 3]);
17. System.out.println(a[99]);
18. }
19. }
20. }
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
BigInteger a[] = new BigInteger[100];
while(cin.hasNext())
{
for(int i = 0; i <= 2; ++i)
a[i] = cin.nextBigInteger();
for(int i = 3; i <= 99; ++i)
a[i] = a[i - 1].add(a[i - 2]).add(a[i - 3]);
System.out.println(a[99]);
}
}
}
高精度幂
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=155
代码如下:
[java] view plaincopyprint?
1. import java.io.*;
2. import java.math.BigDecimal;
3. import java.util.*;
4.
5. public class Main
6. {
7. public static void main(String args[])
8. {
9. Scanner cin = new Scanner(System.in);
10. while(cin.hasNext())
11. {
12. BigDecimal ans = cin.nextBigDecimal();
13. int n = cin.nextInt();
14. String res = ans.pow(n).stripTrailingZeros().toPlainString(); //整数去掉小数点和后面的0
15. if(res.startsWith("0")) //去掉前导0
16. {
17. res = res.substring(1);
18. }
19. System.out.println(res);
20. }
21. }
22. }
计算100!+101!+……+200!
import java.math.*;
public class No0305
{
public static void main(String args[])
{
BigInteger
first100=new BigInteger("100"),//初始化各个大整数
last201=new BigInteger("201"),
one=new BigInteger("1"),
power=new BigInteger("1"),
result=new BigInteger("0");
for (BigInteger i=one;i.compareTo(last201)<0;i=i.add(one))//循环累加从1到201
{
power=power.multiply(i);//求i!
if(i.compareTo(first100)>=0)//从100!开始加
result=result.add(power);
}
System.out.println(result.toString());
}
}
补充:
public static void main(String[] args) {
//int的最大数值是2的31次方减1,如果在最大值上继续增加则会从最小值开始,循环下去。
//BigInteger没有最大值,但计算并不是很方便需要使用里面定义的各种方法才可以。
Integer i = new Integer(2147483647);
BigInteger bi = new BigInteger("2147483647");
System.out.println(i+1);
System.out.println(bi.add(BigInteger.ONE));
}
java中数值型数据的 无穷大、无穷小 的写法
无穷大:
Integer.MAX_VALUE;
Long.MAX_VALUE;
Double.MAX_VALUE
无穷小:
Integer.MIN_VALUE;
Long.MIN_VALUE;
Double.MIN_VALUE;
参考网址:http://www.cnblogs.com/Lyush/category/291376.html
http://blog.csdn.net/niushuai666/article/details/6972991
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