Exponentiation
| Time Limit: 500MS | Memory Limit: 10000K | |
| Total Submissions: 156303 | Accepted: 38063 |
Description
Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.
This problem requires that you write a program to compute the exact value of Rn where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.
Input
The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.
Output
The output will consist of one line for each line of input giving the exact value of R^n. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don‘t print the decimal point if the result is an integer.
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
Hint
If you don‘t know how to determine wheather encounted the end of input:
s is a string and n is an integer
C++
while(cin>>s>>n)
{
...
}
c
while(scanf("%s%d",s,&n)==2) //to see if the scanf read in as many items as you want
/*while(scanf(%s%d",s,&n)!=EOF) //this also work */
{
...
}
Source
East Central North America 1988
原题链接:http://poj.org/problem?id=1001
题意:输入两个数x,y,求x^y,(y为整型),结果小于0,输出的结果去掉前面的"0.",直接输出后面的.
这题和以前的一题差不多,http://blog.csdn.net/hurmishine/article/details/51008733
如:stripTrailingZeros()
toPlainString();
调用的方法也差不多,但这题如果结果小于小于0,要去掉前导"0.",那该怎么做呢?
把结果转换成string类型,再调用java.lang.string类中的substring()方法就OK了.
API中substring()方法的说明:
public String
substring(int beginIndex)
返回一个新的字符串，它是此字符串的一个子字符串。
该子字符串从指定索引处的字符开始，
直到此字符串末尾。
示例：
"unhappy".substring(2) returns "happy"
"Harbison".substring(3) returns "bison"
"emptiness".substring(9) returns "" (an empty string)
参数：
beginIndex - 起始索引（包括）。
返回：
指定的子字符串。
抛出：
IndexOutOfBoundsException - 如果 beginIndex 为负或大于此 String 对象的长度。
substring
public String substring(int beginIndex,int endIndex)
返回一个新字符串，它是此字符串的一个子字符串。
该子字符串从指定的 beginIndex 处开始，直到索引 endIndex - 1 处的字符。
因此，该子字符串的长度为 endIndex-beginIndex。
示例：
"hamburger".substring(4, 8) returns "urge"
"smiles".substring(1, 5) returns "mile"
参数：
beginIndex - 起始索引（包括）。
endIndex - 结束索引（不包括）。
返回：
指定的子字符串。
抛出：
IndexOutOfBoundsException - 如果 beginIndex 为负，或 endIndex 大于此 String 对象的长度，
或 beginIndex 大于 endIndex。
AC代码:
import java.math.BigDecimal;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
BigDecimal a = sc.nextBigDecimal();
int b = sc.nextInt();
a = a.pow(b);
String str = a.stripTrailingZeros().toPlainString();
if (str.startsWith("0.")){
str = str.substring(1);
}
System.out.println(str);
}
}
}