Exponentiation
| Time Limit: 500MS | Memory Limit: 10000K | |
| Total Submissions: 190785 | Accepted: 45785 |
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
這個數字，那麼大，哪怕是longlonglonglong來個十個八個的也不夠用，那這樣的，用java語言自帶的+-*/那肯定不行，如果行的話考這個也沒什麼意思了。
所以只能用數組存儲。
其實吧，我覺得，要和小學的豎式一樣，就無敵了。
核心思想是：
假如兩個數相乘
圖有點亂，大意就是，int a[] 存儲第一個數組，每個位存一個數。int b[]第二個
用int c[]，存儲a * b 的值
那麼就會發現一個規律，c[1] 是不是等於a[1] * b[1].
c[2]是不是等於a[1]*b[2]+a[2]*b[1].看我在圖上畫的線，c[3] = a[3]*a[1] + a[1]*b[3];
這樣其實就可以把這個值存儲起來了
但是一想
着c存的啥呀，這不是162 162 .。。。。
這怎麼輸出呀。
不用着急，咱們豎式不就滿十進一嘛。
如果當前是81，那應該進8個，自己留一個。
也就是說，一個循環判斷，從低往高位，滿十進一。
c[i+1] += c[i]/10.
c[i] %10;
就行了。這就是算法的核心。
那a[0]位存啥，一個蘿蔔一個坑，0位拿來存小數點的位置。
import java.util.Scanner;
class BigNumber{
int[] result;
int[] value;
int len;
BigNumber(int[] result,int[] value){
this.result = result;
this.value = value;
}
public int[] getResult() {
return result;
}
public void setResult(int[] result) {
this.result = result;
}
}
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String s;
while(sc.hasNext()) {
int[] a = new int[100];
s = sc.next();
//6位的小數
for(int i=s.length()-1,k=0;i>=0;i--) {
if(Character.isDigit(s.charAt(i))) {
a[++k] = s.charAt(i) - '0';
}else if(s.charAt(i) == '.') {
a[0] = k;
}
}
int[] b = filter(a);
int power = sc.nextInt();
sc.nextLine();
BigNumber bg = new BigNumber(b,b);
while(true) {
power--;
if(power == 0) break;
bg.setResult(mulp(bg.result,bg.value));
}
int[] result = filter(bg.getResult());
int len = getLen(result);
power = result[0];
if(power > len ) {
System.out.print(".");
for(int i=0;i<power-len;i++) {
System.out.print("0");
}
}
for(int i=len,count=0;i>=1;i--) {
System.out.print(result[i]);
if(++count == len-power && power != 0) System.out.print(".");
}
if(getLen(result) == 0 && result[0] == 0) {
System.out.print("0");
}
System.out.print("\n");
}
}
public static int[] mulp(int[] a,int[] b) {
int[] c = new int[100];
c[0] = a[0]+b[0];
for(int i=1;i<=getLen(a);i++) {
for(int j=1;j<=getLen(b);j++) {
c[i+j-1] += a[i]*b[j];
}
}
for(int i=1;i<=getLen(c);i++) {
if(c[i]>9) {
c[i+1] += c[i]/10;
c[i] %= 10;
}
}
return c;
}
public static int getLen(int[] a) {
int i = a.length-1;
while(a[i] == 0 && i > 0) {
i--;
}
return i;
}
public static int[] filter(int[] a) {
int k = 0;
for(int i=1;i<=getLen(a);i++) {
if(a[i] == 0) {
if(a[0] == 0) break;
a[0]--;
k++;
}else {
break;
}
}
if(k != 0) {
int l = getLen(a);
for(int i=k+1;i<=l;i++) {
a[i-k] = a[i];
}
for(int i=l;i>l-k;i--) {
a[i] = 0;
}
}
if(getLen(a) == 0) {
a[0] = 0;
}
return a;
}
}
思路是很簡單的，但是這裏可能會有幾個坑點。
多測幾組數據
0.0000 2
1.0000 2
0.0001 2