顺序我正在检查输出错误的字符串
Order I am checking strings getting wrong output
我是c++的新手,虽然我有Java的基本知识,但我正在尝试制作一个程序,转换罗马数字输入,然后找到等效的阿拉伯数字并输出。然而,我遇到了一个问题,即如何在罗马数字中找到特定前缀的正确顺序,例如,因为它要单独检查"X"并删除它;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查找前缀的顺序有关,但我不确定如何修复它,所以在这方面提供任何帮助都会很好。
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;
}
如有任何帮助,我们将不胜感激。
使用贪婪解析策略。也就是说,只要你有一个有效的数字,就可以消耗尽可能多的输入。例如,使用您的示例输入:
MCCCXXXVII
首先看M
,它是一个有效的数字。然后看MC
,这不是一个有效的数字,所以你消耗M,然后从C
开始。
C
是有效数字吗?对CC
是有效数字吗?对CCC
是有效数字吗?对CCCX
是有效数字吗?否,因此消耗CCC
,然后从X
重新开始。
看到照片了吗?
我没有多想,但在我看来,这个问题比你试图解决的问题更简单……我相信你应该能够一次一个字符地解析字符串,更新值。
一般的想法是,从结果设置为0开始,一次读取一个字符,同时记住前一个字符(最初设置为无法出现在输入中的字符。对于读取的每个字符,将该值乘以该字符的值(M=1000,D=500,C=100,…)
当读取的最后一个字符的面额低于此字符时,就会出现奇怪的情况。在这种情况下,您只需要修复该值。
例如,在MCD中,你会开始递增1000,然后递增100,然后找到500,但这比C大,所以你需要修正:递减之前添加的100,加400,总共1400。
您应该使用调试器!以下是我的发现:
-
IX
中的用户类型 -
一切都很好,直到你进入贯穿你的
tens
阵列的for
循环: -
在
tens
数组中循环时,会检查else if (line.find("X") != string::npos)
-
返回
true
。应该这样吗?9
不应在tens
中。 -
一个问题的解决方案是检查它是否不是
9
:else if (line.find("X") != string::npos && line != "IX")
编辑:
这只是一个快速的破解,从长远来看,你应该听从David Rodríguez的建议,简化你的解决方案。
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