c - 贪心算法 - 罗马数字

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我正在尝试实现一个转换整数 n 的 C 程序(0 到 1024 之间)通过贪心算法转换为罗马数字。我尝试通过以下方式做到这一点:

#include <stdio.h>
#include <string.h>
void convert(int);
int max(int[], int, int);

int main(){

    //User Input
    int n;
    printf("Enter a digit between 0 and 1024: ");
    scanf("%d", &n);

    //Validation
    while((n < 0)||(n > 1024)){
        printf("That number is not between 0 and 1024. Please try again: ");
        scanf("%d", &n);
    }

    //Output
    printf("\nAs a Roman numeral, this was written: ");
    if (n == 0) printf("nulla");    //Romans wrote 'nulla' for zero
    else convert(n);

    return 0;
}

void convert(int n){
    //Case n = 0
    if (n == 0)  return;

    else{
        //Case n > 0
        char* romanNums[] = {"I", "IV", "V", "IX", "X", "XL", "L", "XC", "C", "CD", "D", "CM", "M"};
        int arabicNums[] = {1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000};
        int biggestNo = max(arabicNums, 12, n); //Biggest number in arabicNums[] smaller than n

        printf("%s", romanNums[biggestNo]); //Print the biggest number as a Roman numeral
        convert(n - biggestNo);             //Convert the remaining

    }
}

//This function determines the maximum number in arr[] smaller than n
int max(int arr[], int size, int n){
    int i, max;
    for(i = 0; i < size; i++){
        if (n < arr[i]) max = i;
    }
    return max;
}

我尝试过调试和修改代码的各个方面，但它不起作用。如果有任何反馈，我将不胜感激。

<小时/>

更新 我已设法修改我的程序，使其正确输出值 1、4、5 等，但复合值(即需要再次迭代 convert() 的值)保留导致“Romans.exe 没有响应”。这是新代码:

#include <stdio.h>
#include <string.h>
void convert(int);
int max(int[], int, int);

int main(){

    //User Input
    int n;
    printf("Enter a digit between 0 and 1024: ");
    scanf("%d", &n);

    //Validation
    while((n < 0)||(n > 1024)){
        printf("That number is not between 0 and 1024. Please try again: ");
        scanf("%d", &n);
    }

    //Output
    printf("\nAs a Roman numeral, this was written: ");
    if (n == 0) printf("nulla");    //Romans wrote 'nulla' for zero
    else convert(n);

    return 0;
}

void convert(int n){
    //Case n = 0
    if (n == 0)  return;

    else{
        //Case n > 0
        char* romanNums[] = {"I", "IV", "V", "IX", "X", "XL", "L", "XC", "C", "CD", "D", "CM", "M"};
        int arabicNums[] = {1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000};
        int biggestNo = max(arabicNums, 13, n); //Biggest number in arabicNums[] smaller than n

        printf("%s", romanNums[biggestNo]); //Print the biggest number as a Roman numeral
        convert(n - arabicNums[biggestNo]); //Convert the remaining

    }
}

//This function determines the maximum number in arr[] smaller than n
int max(int arr[], int size, int n){
    int i, max;
    for(i = size; i > 0; i--){
        if (n <= arr[i]) max = i;
    }
    return max;
}

Answer 1

您的代码中有 2 个缺失点:

最大功能:

for(i = 0; i < size; i++){
    if (n < arr[i]) max = i;
}

必须是:

for(i = 0; i <= size; i++){
    if (n >= arr[i]) max = i;   // Equal is require. Isn't it?
}

在您的主要功能上:函数转换:

convert(n - biggestNo);             //Convert the remaining

必须是:

convert(n - arabicNums[biggestNo]);

biggestNo 是 seq 编号，而不是 div 的值