我的作业是写一个简单的程序，将数字转换成罗马数字。我可以使用的唯一功能是printf和scanf。我已经编写了代码，但是我不知道如何打印出诸如＆＃34; 25,13,93,66和＃34;等任何帮助将不胜感激！谢谢！ （以下是书面代码）
int main(int arg, char *argv[]){
int thousand, hundred, ten, single;
int number;
printf("Enter a decimal number: ");
scanf("%d", &number);
thousand = number + 0;
if (thousand == 1000) {
printf("%d is represented as M in Roman numerals. \n", number);
} else if (thousand == 2000) {
printf("%d is represented as MM in Roman numerals. \n", number);
} else if (thousand == 3000)
printf("%d is represented as MMM in Roman numerals. \n",number);
hundred = number + 0;
if (hundred == 100) {
printf("%d is represented as C in Roman numerals. \n",number);
} else if (hundred == 200) {
printf("%d is represented as CC in Roman numerals. \n",number);
} else if (hundred == 300) {
printf("%d is represented as CCC in Roman numerals. \n",number);
} else if (hundred == 400) {
printf("%d is represented as CD in Roman numerals. \n",number);
} else if (hundred == 500) {
printf("%d is represented as D in Roman numerals. \n",number);
} else if (hundred == 600) {
printf("%d is represented as DC in Roman numerals. \n",number);
} else if (hundred == 700) {
printf("%d is represented as DCC in Roman numerals. \n",number);
} else if (hundred == 800) {
printf("%d is represented as DCCC in Roman numerals. \n",number);
} else if (hundred == 900)
printf("%d is represented as CM in Roman numerals. \n",number);
ten = number + 0;
if (ten == 10) {
printf("%d is represented as X in Roman numerals. \n",number);
} else if (ten == 20) {
printf("%d is represented as XX in Roman numerals. \n",number);
} else if (ten == 30) {
printf("%d is represented as XXX in Roman numerals. \n",number);
} else if (ten == 40) {
printf("%d is represented as XL in Roman numerals. \n",number);
} else if (ten == 50) {
printf("%d is represented as L in Roman numerals. \n",number);
} else if (ten == 60) {
printf("%d is represented as LX in Roman numerals. \n",number);
} else if (ten == 70) {
printf("%d is represented as LXX in Roman numerals. \n",number);
} else if (ten ==80) {
printf("%d is represented as LXXX in Roman numerals. \n",number);
} else if (ten == 90)
printf("%d is represented as XC in Roman numerals. \n",number);
single = number + 0;
if (single == 1) {
printf("%d is represented as I in Roman numerals. \n",number);
} else if (single == 2) {
printf("%d is represented as II in Roman numerals. \n",number);
} else if (single == 3) {
printf("%d is represented as III in Roman numerals. \n",number);
} else if (single == 4) {
printf("%d is represented as IV in Roman numerals. \n",number);
} else if (single == 5) {
printf("%d is represented as V in Roman numerals. \n",number);
} else if (single == 6) {
printf("%d is represented as VI in Roman numerals. \n",number);
} else if (single == 7) {
printf("%d is represented as VII in Roman numerals. \n",number);
} else if (single ==8) {
printf("%d is represented as VIII in Roman numerals. \n",number);
} else if (single == 9)
printf("%d is represented as IX in Roman numerals. \n",number);
if (number <= 0) {
printf("%d can not be represented in Roman numerals. \n", number);
} else if (number > 3000)
printf("%d can not be represented in Roman numerals. \n", number);
return 0;
}
答案 0 :(得分：1)
您似乎尝试使用以下内容连接数字 百=数字+ 0;
这只会导致数字，数字+ 0 =数字。如果要添加零，请将数字乘以10.如下所示 百=数字* 10;
对于软件设计，如果程序无法转换数字，您可能希望返回非零数字，因为这可以通过其他调用程序/脚本进行分析。
我发现在使用缩进格式正确格式化代码时更容易查看代码，因为这样可以更全面地了解代码流以及范围落在何处。
我希望这些小技巧可以帮助:)。
答案 1 :(得分：1)
void dec2romanstr(int num){
int del[] = {1000,900,500,400,100,90,50,40,10,9,5,4,1}; // Key value in Roman counting
char * sym[] = { "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" }; //Symbols for key values
char res[64] = "\0"; //result string
int i = 0; //
while (num){ //while input number is not zero
while (num/del[i]){ //while a number contains the largest key value possible
strcat(res, sym[i]); //append the symbol for this key value to res string
num -= del[i]; //subtract the key value from number
}
i++; //proceed to the next key value
}
puts(res);
}
答案 2 :(得分：0)
#include <stdio.h>
#include <stdlib.h>
typedef struct radix {
int digit;
char roman;
char romanNext;
} RADIX;
char *roman(int num, char *buff){
static RADIX table[]={
{ 1000, 'M', '?' },
{ 100, 'C', 'D' },
{ 10, 'X', 'L' },
{ 1, 'I', 'V' }
};
int tableSize = sizeof(table)/sizeof(RADIX);
int i, j, n;
char *p;
if(num < 0 || num >=4000){
*buff='\0';
return NULL;
}
for(i=0,p=buff;i<tableSize;i++){
n = num / table[i].digit;
if( 1 <= n && n <=3 ){
for(j=0;j<n;j++)
*p++=table[i].roman;
} else if(n == 4){
*p++=table[i].roman;
*p++=table[i].romanNext;
} else if(n == 5){
*p++=table[i].romanNext;
} else if(6<= n && n <=8){
*p++=table[i].romanNext;
for(j=0;j< n-5;++j)
*p++=table[i].roman;
} else if(n == 9){
*p++=table[i].roman;
*p++=table[i-1].roman;
}
num -= n * table[i].digit;
}
*p='\0';
return buff;
}
int main(){
int i, number[] = {25, 13, 93, 66};
char buff[16];
for(i=0;i < 4;++i){
if(roman(number[i], buff))
printf("%d is %s\n", number[i], buff);
}
return 0;
}
答案 3 :(得分：0)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct {
int value;
char romanDgts[3];
} romanTable[] = {{1000, "M"},
{900, "CM"},
{500, "D"},
{400, "CD"},
{100, "C"},
{90, "XC"},
{50, "L"},
{40, "XL"},
{10, "X"},
{9, "IX"},
{5, "V"},
{4, "IV"},
{1, "I"}};
char *intToRoman(int n)
{
char *romanNumerals;
int i = 0;
romanNumerals = (char *)malloc(sizeof(char) * 10);
while (n) {
while (n < romanTable[i].value)
i++;
while(n >= romanTable[i].value) {
strcat(romanNumerals, romanTable[i].romanDgts);
n -= romanTable[i].value;
}
}
return romanNumerals;
}
int main(int argc, char const *argv[])
{
int number;
char *result;
printf("Input a decimal number: ");
scanf("%d", &number);
if(number < 0 || number > 3999) {
printf("Input must be within the range from 1 to 3999.\n");
return -1;
}
result = intToRoman(number);
puts(result);
free(result);
return 0;
}
答案 4 :(得分：0)
#include<stdio.h>
int main(){
int num,reminder,counter;
scanf("%d",&num);
for(counter=1;counter<=num/1000;counter++)
printf("M");
reminder = num%1000;
for(counter=reminder/900;counter==1;counter++)
printf("CM");
reminder = reminder%900;
for(counter=reminder/500;counter==1;counter++)
printf("D");
reminder = reminder%500;
for(counter=reminder/400;counter==1;counter++)
printf("CD");
reminder = reminder%400;
for(counter=1;counter<=reminder/100;counter++)
printf("C");
reminder = reminder%100;
for(counter=reminder/90;counter==1;counter++)
printf("XC");
reminder = reminder%90;
for(counter=reminder/50;counter==1;counter++)
printf("L");
reminder = reminder%50;
for(counter=reminder/40;counter==1;counter++)
printf("XL");
reminder = reminder%40;
for(counter=reminder/10;counter>=1;counter--)
printf("X");
reminder = reminder%10;
for(counter=reminder/9;counter==1;counter++)
printf("IX");
reminder = reminder%9;
for(counter=reminder/5;counter==1;counter++)
printf("V");
reminder = reminder%5;
for(counter=reminder/4;counter==1;counter++)
printf("IV");
reminder = reminder%4;
for(counter=reminder;counter>=1;counter--)
printf("I");
}
长（43行），但简单对!! !!
答案 5 :(得分：-1)
问题确实令人生畏，至少在初学者水平上......因为你提到只使用基本功能，所以使用我的解决方案可以轻松打印 1000 以下的数字：< / p>
# include<stdio.h>
main()
{
int num,r,t,z,count=1;
char c1='I',c2='V',c3='X',c4='L',c5='C',c6='D',c7='M';
printf("enter the number");
scanf("%d",&num);
r=num%10;
z=num/100;
t=num%100;
while(count<=3)
{
if(count==2)
{
z=(t-r)/10;
c7='C';
c6='L';
c5='X';
}
if(count==3)
{
z=r;
c7='X';
c6='V';
c5='I';
}
switch(z)
{
case 0:printf("");
break;
case 1:printf("%c",c5);
break;
case 2:printf("%c%c",c5,c5);
break;
case 3:printf("%c%c%c",c5,c5,c5);
break;
case 4:printf("%c%c",c5,c6);
break;
case 5:printf("%c",c6);
break;
case 6:printf("%c%c",c6,c5);
break;
case 7:printf("%c%c%c",c6,c5,c5);
break;
case 8:printf("%c%c%c%c",c6,c5,c5,c5);
break;
case 9:printf("%c%c",c5,c7);
break;
case 10:printf("%c",c7);
break;
}
count++;
}
}