[UVA][Java] 288 - Arithmetic Operations With Large Integers
| Arithmetic Operations With Large Integers |
This problem is about calculations with large numbers. Large means numbers with at most one thousand digits. The operations are limited to addition, substraction, multiplication and raising to a higher power.
There are no limitations to the operants of addition, substraction and multiplication. The base in raising to higher powers is positive and smaller than ten. The exponent is positive.
The inputfile consists of a valid expression with any number of operations. There are no parentheses, but the normal arithmetic priority rules are still valid.
An example of a valid expression is: 12345678 * 129876 + 2**1993. An invalid expression is: 12345678 * 129876 + 12**1993 because the base is greater than nine.
Input
The input contains several test cases, each one on a different line. Each test case contains numbers and operands in the following way: n op n { op n } . n is a positive decimal number with at most one thousand digits, stored as an ASCII-text. op is one of the following: +, -, *, ** (** means ``raising to higher powers''). There can be at most one hundred operations per test case. There are no spaces or other illegal characters in the input.
Output
The output contains the exact result of the evaluated expressions given in the input. Print each test case in a different line. Each test case won't have more than three thousand characters.
Sample Input
12345678*129876+2**1993
Sample Output
896977105683011347056900938420064050017435704756793125373158388145129891712\\ 789307700515223684770523373785909874208955291755561688174261977676508872005\\ 197801086953040197752187505381087095625350558038492109870986287356370809737\\ 409093338414265941143390397695285610643740694879918793932122262001282984143\\ 224073001319601441082075018589725061828585163552941409601583724270514300953\\ 188533095947591884905338415676554651534516617357655143781579373852994152663\\ 198702360093129335607684294312805938140290754926427776409574872859496315224\\ 893901812925850900592061583009183090068756428459147015355107518672556877720
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
String line;
while(cin.hasNextLine()) {
line = cin.nextLine();
BigInteger[] S = new BigInteger[201];
int[] neg = new int[200];
int Idx = 0, len = line.length();
int oper = 0;
for(int i = 0; i < len; i++) {
String sub = "";
while(i < len && Character.isDigit(line.charAt(i))) {
sub += line.charAt(i);
i++;
}
if(oper == 0) { // add
S[Idx] = new BigInteger(sub);
neg[Idx] = 0;
Idx++;
} else if(oper == 1) {// subtract
S[Idx] = new BigInteger(sub);
neg[Idx] = 1;
Idx++;
} else if(oper == 2) {// multiply
S[Idx-1] = S[Idx-1].multiply(new BigInteger(sub));
} else {
if(S[Idx-1].longValue() < 2)
continue;
S[Idx-1] = S[Idx-1].pow(Integer.parseInt(sub));
}
if(i < len && line.charAt(i) == '+') {
oper = 0;
} else if(i < len && line.charAt(i) == '-') {
oper = 1;
} else if(i < len && line.charAt(i) == '*') {
if(i+1 < len && line.charAt(i+1) == '*') {
i++;
oper = 3;
} else {
oper = 2;
}
}
}
BigInteger ans = new BigInteger("0");
for(int i = 0; i < Idx; i++) {
if(neg[i] == 1)
ans = ans.subtract(S[i]);
else
ans = ans.add(S[i]);
}
System.out.println(ans);
}
}
}
請問這題如果用C++寫要怎麼做呢
還有判斷先乘除 指數後加減的問題在上面的code是在哪個部分
例如 1 * 2 - 3 * 4 => 最後加總 2 - 12。
讀入的過程中
=> 1
=> 2
=> 2 -3
=> 2 -12
最後加總所有數字即可。如果遇到乘除法，依附在前一個數字即可，反之斷開成為一個新的數字。 2014-11-01 18:18:54