Question

如果自然数小于或等于99，则将其打印为罗马数字。

有这方法吗？我该如何解决这个任务？ Noob在这里提问。

Answer 1

这段代码可行，应该是自我解释的。

#include <stdio.h>

void printRomanNumerals(int input);

int main(int argc, const char * argv[]) {
    printf("Input your number!\n");
    int i;
    scanf("%d", &i);
    printRomanNumerals(i);
    return 0;
}

void printRomanNumerals(int input) {

    switch (input / 10) {
        case 1:
            printf("%s","X");
            break;
        case 2:
            printf("%s","XX");
            break;
        case 3:
            printf("%s","xxx");
            break;
        case 4:
            printf("%s","XL");
            break;
        case 5:
            printf("%s","L");
            break;
        case 6:
            printf("%s","LX");
            break;
        case 7:
            printf("%s","LXX");
            break;
        case 8:
            printf("%s","LXXX");
            break;
        case 9:
            printf("%s","XC");
            break;
        default:
            break;
    }

    switch (input % 10) {
        case 1:
            printf("%s","I");
            break;
        case 2:
            printf("%s","II");
            break;
        case 3:
            printf("%s","III");
            break;
        case 4:
            printf("%s","IV");
            break;
        case 5:
            printf("%s","V");
            break;
        case 6:
            printf("%s","VI");
            break;
        case 7:
            printf("%s","VII");
            break;
        case 8:
            printf("%s","VII");
            break;
        case 9:
            printf("%s","IX");
            break;
        default:
            break;
    }

}

Answer 2

这些链接页面上的一些答案太复杂了。为什么新手会得到所有有趣的问题......

#include <stdio.h>

const char *digits[] = { "", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX" };
const char *prefixes[] = { "", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC" };

#define LIMIT sizeof(digits) / sizeof(char *) * sizeof(prefixes) / sizeof(char *)

int main() {
    int i = 0;

    do {
        printf("Da mihi numeralis (0 si velis relinquere): ");
        (void) scanf("%2i", &i);
        (void) fpurge(stdin);
        if (0 < i && i < LIMIT) {
            printf("%s%s\n", prefixes[i / 10], digits[i % 10]);
        }
    } while (i > 0);

    return 0;
}

C - 如何打印罗马数字？

2 个答案: