虽然我有Java的基本知识，但我是c++新手，我正在尝试制作一个程序来转换罗马数字输入，然后找到等效的阿拉伯数字并输出它。然而，我遇到了一个问题，例如，如何在罗马数字中查找特定前缀时找到正确的顺序，因为它单独检查“X”并将其删除；我的程序将输入“IX”翻译为 11，而实际上它是 9。我知道这与我的程序在字符串中找到前缀的顺序有关，但是我不确定如何修复它，所以任何对此提供帮助就太好了。
当前代码:
#include <cstdlib>
#include <iostream>
#include <cctype>
using namespace std;
/*
*
*/
int main() {
string roman_digits [] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
string roman_tens [] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
string roman_hundreds [] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
string roman_thousands [] = {"", "M", "MM", "MMM"};
string line, substr;
int arabic = 0;
// MCCCXXXVII
cout << "Type in a Roman numeral: ";
// Loops through inputted Roman Numerals.
while (cin >> line) {
if (!cin.eof()) {
int i = 0;
// Loops through a Roman numeral and changes it to uppercase.
while (line[i]) {
char c;
c = line[i];
c = (toupper(c));
line[i] = c;
i++;
}
// Loops through checking roman numeral with the thousands array and if there is a match prints out the equivalent arabic number.
for (int i = 0; i < 4; i++) {
if (line.find("MMM") != string::npos) {
unsigned pos = line.find("MMM");
substr = line.substr(pos, 3);
line.erase(pos, 3);
} else if (line.find("MM") != string::npos) {
unsigned pos = line.find("MM");
substr = line.substr(pos, 2);
line.erase(pos, 2);
} else if (line.find("M") != string::npos) {
unsigned pos = line.find("M");
substr = line.substr(pos, 1);
line.erase(pos, 1);
}
if (roman_thousands[i] == substr){
arabic = arabic + (i * 1000);
}
}
// Loops through checking roman numeral with the hundreds array and if there is a match prints out the equivalent arabic number.
for (int i = 0; i < 10; i++) {
if (line.find("CM") != string::npos){
unsigned pos = line.find("CM");
substr = line.substr(pos, 2);
line.erase(pos, 2);
} else if (line.find("DCCC") != string::npos){
unsigned pos = line.find("DCCC");
substr = line.substr(pos, 4);
line.erase(pos, 4);
} else if (line.find("DCC") != string::npos){
unsigned pos = line.find("DCC");
substr = line.substr(pos, 3);
line.erase(pos, 3);
} else if (line.find("DC") != string::npos){
unsigned pos = line.find("DC");
substr = line.substr(pos, 2);
line.erase(pos, 2);
} else if (line.find("D") != string::npos){
unsigned pos = line.find("D");
substr = line.substr(pos, 1);
line.erase(pos, 1);
} else if (line.find("CD") != string::npos){
unsigned pos = line.find("CD");
substr = line.substr(pos, 2);
line.erase(pos, 2);
} else if (line.find("CCC") != string::npos){
unsigned pos = line.find("CCC");
substr = line.substr(pos, 3);
line.erase(pos, 3);
}else if (line.find("CC") != string::npos){
unsigned pos = line.find("CC");
substr = line.substr(pos, 2);
line.erase(pos, 2);
} else if (line.find("C") != string::npos){
unsigned pos = line.find("C");
substr = line.substr(pos, 1);
line.erase(pos, 1);
}
if (roman_hundreds[i] == substr) {
arabic = arabic + (i * 100);
}
}
// Loops through checking roman numeral with the tens array and if there is a match prints out the equivalent arabic number.
for (int i = 0; i < 10; i++) {
if (line.find("XC") != string::npos){
unsigned pos = line.find("XC");
substr = line.substr(pos, 2);
line.erase(pos, 2);
} else if (line.find("LXXX") != string::npos){
unsigned pos = line.find("LXXX");
substr = line.substr(pos, 4);
line.erase(pos, 4);
}else if (line.find("LXX") != string::npos){
unsigned pos = line.find("LXX");
substr = line.substr(pos, 3);
line.erase(pos, 3);
} else if (line.find("LX") != string::npos){
unsigned pos = line.find("LX");
substr = line.substr(pos, 2);
line.erase(pos, 2);
}else if (line.find("L") != string::npos){
unsigned pos = line.find("L");
substr = line.substr(pos, 1);
line.erase(pos, 1);
}else if (line.find("XL") != string::npos){
unsigned pos = line.find("XL");
substr = line.substr(pos, 2);
line.erase(pos, 2);
}else if (line.find("XXX") != string::npos){
unsigned pos = line.find("XXX");
substr = line.substr(pos, 3);
line.erase(pos, 3);
}else if (line.find("XX") != string::npos){
unsigned pos = line.find("XX");
substr = line.substr(pos, 2);
line.erase(pos, 2);
}else if (line.find("X") != string::npos){
unsigned pos = line.find("X");
substr = line.substr(pos, 1);
line.erase(pos, 1);
}
if (roman_tens[i] == substr) {
arabic = arabic + (i * 10);
}
}
// Loops through checking roman numeral with the digits array and if there is a match prints out the equivalent arabic number.
for (int i = 0; i < 10; i++) {
if (line.find("IX") != string::npos){
unsigned pos = line.find("IX");
substr = line.substr(pos, 2);
line.erase(pos, 2);
} else if (line.find("VIII") != string::npos){
unsigned pos = line.find("VIII");
substr = line.substr(pos, 4);
line.erase(pos, 4);
} else if (line.find("VII") != string::npos){
unsigned pos = line.find("VII");
substr = line.substr(pos, 3);
line.erase(pos, 3);
} else if (line.find("VI") != string::npos){
unsigned pos = line.find("VI");
substr = line.substr(pos, 2);
line.erase(pos, 2);
} else if (line.find("V") != string::npos){
unsigned pos = line.find("V");
substr = line.substr(pos, 1);
line.erase(pos, 1);
} else if (line.find("IV") != string::npos){
unsigned pos = line.find("IV");
substr = line.substr(pos, 2);
line.erase(pos, 2);
} else if (line.find("III") != string::npos){
unsigned pos = line.find("III");
substr = line.substr(pos, 3);
line.erase(pos, 3);
} else if (line.find("II") != string::npos){
unsigned pos = line.find("II");
substr = line.substr(pos, 2);
line.erase(pos, 2);
}else if (line.find("I") != string::npos){
unsigned pos = line.find("I");
substr = line.substr(pos, 1);
}
if (roman_digits[i] == substr) {
arabic = arabic + i;
}
}
cout << "The Arabic equivalent of " << line << " is: " << arabic << endl;
arabic = 0;
} else {
cout << "Invalid Roman numeral." << endl;
}
}
return 0;
}
任何帮助将不胜感激，谢谢。
最佳答案
使用贪心解析策略。也就是说，只要你有一个有效的数字，就消耗尽可能多的输入。例如，使用您的示例输入:
MCCCXXXVII
首先查看M，这是一个有效的数字。然后您查看 MC，它不是有效数字，因此您使用 M，然后从 C 重新开始。
C 是有效数字吗？是的。 CC 是有效数字吗？是的。 CCC 是有效数字吗？是的。 CCCX 是有效数字吗？不，所以消耗 CCC 并在 X 处重新开始。
有照片吗？
关于c++ - 命令我检查字符串得到错误的输出，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/18524156/