int型数字转化为罗马数字的形式
思路：
由于只是1到3999,一共只有四位，分别求这四位的情况。
可以将每一位从1到9，int和罗马数字的一一对应的关系给出来，然后直接转换。
/*****************************************************************
Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.
*****************************************************************/
#include <stdio.h>
char* intToRoman(int num) {
int sd = num %10;
int nm = num/10;
int td = nm%10;
nm /= 10;
int hd = nm%10;
nm /= 10;
int thd = nm%10;
int i,index = 0;
char *rstr = (char *)malloc(16*sizeof(char));
memset(rstr,0,16*sizeof(char));
for(i = 0;i < thd;i++){
rstr[index++] = ‘M‘;
}
switch (hd)//百位
{
case 3://注意没有break
rstr[index++] = ‘C‘;
case 2://注意没有break
rstr[index++] = ‘C‘;
case 1://注意没有break
rstr[index++] = ‘C‘;
break;
case 4://注意没有break
rstr[index++] = ‘C‘;
case 5:
rstr[index++] = ‘D‘;
break;
case 6:
rstr[index++] = ‘D‘;
rstr[index++] = ‘C‘;
break;
case 7:
rstr[index++] = ‘D‘;
rstr[index++] = ‘C‘;
rstr[index++] = ‘C‘;
break;
case 8:
rstr[index++] = ‘D‘;
rstr[index++] = ‘C‘;
rstr[index++] = ‘C‘;
rstr[index++] = ‘C‘;
break;
case 9:
rstr[index++] = ‘C‘;
rstr[index++] = ‘M‘;
break;
default:
break;
}
switch (td)//十位
{
case 3:
rstr[index++] = ‘X‘;
case 2:
rstr[index++] = ‘X‘;
case 1:
rstr[index++] = ‘X‘;
break;
case 4:
rstr[index++] = ‘X‘;
case 5:
rstr[index++] = ‘L‘;
break;
case 6:
rstr[index++] = ‘L‘;
rstr[index++] = ‘X‘;
break;
case 7:
rstr[index++] = ‘L‘;
rstr[index++] = ‘X‘;
rstr[index++] = ‘X‘;
break;
case 8:
rstr[index++] = ‘L‘;
rstr[index++] = ‘X‘;
rstr[index++] = ‘X‘;
rstr[index++] = ‘X‘;
break;
case 9:
rstr[index++] = ‘X‘;
rstr[index++] = ‘C‘;
break;
default:
break;
}
switch (sd)//个位
{
case 3:
rstr[index++] = ‘I‘;
case 2:
rstr[index++] = ‘I‘;
case 1:
rstr[index++] = ‘I‘;
break;
case 4:
rstr[index++] = ‘I‘;
case 5:
rstr[index++] = ‘V‘;
break;
case 6:
rstr[index++] = ‘V‘;
rstr[index++] = ‘I‘;
break;
case 7:
rstr[index++] = ‘V‘;
rstr[index++] = ‘I‘;
rstr[index++] = ‘I‘;
break;
case 8:
rstr[index++] = ‘V‘;
rstr[index++] = ‘I‘;
rstr[index++] = ‘I‘;
rstr[index++] = ‘I‘;
break;
case 9:
rstr[index++] = ‘I‘;
rstr[index++] = ‘X‘;
break;
default:
break;
}
return rstr;
}
int main(){
int n = 1684;
char *s = intToRoman(n);
printf("%s\n",s);
free(s);
return 0;
}
上面思路代码冗余，不好。
思路2：
考虑每位对应数字是多少就先填多少个对应的罗马数字；
例如：46->XXXXIIIIII
然后多余的罗马数字在合。
合并的可能情况如下：
注意，每位都是一样的，用个位举例。
1.4个I合并为IV；IIII->IV
2.5个I合并为V；IIIII->V
3.2个V合并为X；IIIIIIIII->VIIII->VIV->IX
1 /*****************************************************************
2 Given an integer, convert it to a roman numeral.
3 Input is guaranteed to be within the range from 1 to 3999.
4 *****************************************************************/
5 #include <stdio.h>
6
7 char* intToRoman(int num) {
8 char *rstr = (char *)malloc(32*sizeof(char));
9 memset(rstr,0,32*sizeof(char));
10 char lromans[] = {‘M‘,‘C‘,‘X‘,‘I‘};
11 char uromans[] = {‘ ‘,‘D‘,‘L‘,‘V‘};
12 int count = 0,lcount = 0,up = 0;
13 int i,k = 1000,index = 0,sum = 0;
14 while(num > 0){//在rstr数组中填充num的每位的数值对应数量的单位量字母（如num=9，则rstr="IIIIIIIII"）
15 up = num/k;
16 num = num%k;
17 for(i = 0;i < up;i++){
18 rstr[sum++] = lromans[count];//填充
19 }
20 count++;
21 k = k/10;
22 }
23
24 count = 0;
25 for(i = 0;i < sum;i++){//将每位的数值对应数量的单位量字母转换成正确的表示法
26 rstr[index] = rstr[i];
27 if(rstr[i] == lromans[count]){
28 lcount++;
29 }else{
30 do{
31 count++;
32 }while(rstr[i] != lromans[count]);
33 lcount = 1;
34 }
35 if(lcount == 4){//4个的情况：IIII->IV
36 index -= 2;
37 rstr[index] = uromans[count];
38 }else if(lcount == 5){//5个的情况：IIIII->V
39 index -= 2;
40 rstr[index] = uromans[count];
41 }else if(lcount == 9){//9个的情况：IIIIIIIII->VIIII->VIV->IX
42 index -= 4;
43 rstr[index++] = lromans[count];
44 rstr[index] = lromans[count - 1];
45 }
46 index++;
47 }
48 rstr[index] = ‘\0‘;
49
50 return rstr;
51 }
52
53 int main(){
54 int n = 3999;
55 char *s = intToRoman(n);
56 printf("%s\n",s);
57 free(s);
58 return 0;
59 }
思路3：
贪心算法：尽量从大单位开始匹配，但是4和9要特殊考虑。
/*****************************************************************
Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.
*****************************************************************/
#include <stdio.h>
//找钱的思想，对4和9特殊处理，贪心算法
char* intToRoman(int num) {
int weights[] = {1000,500,100,50,10,5,1};
char romans[] = {‘M‘,‘D‘,‘C‘,‘L‘,‘X‘,‘V‘,‘I‘};
int nums[7] = {0};
char *rstr = (char *)malloc(16*sizeof(char));
memset(rstr,0,16*sizeof(char));
int i = 0,j = 0,index = 0;
while(num > 0){
nums[i] = num/weights[i];//记录罗马数字每个单位对应的字母的个数
num = num%weights[i];
i++;
}
for(i = 0;i < 7;i++){
if(nums[i] == 4){//低位的单位有4个的时候可以转化的高位单位：IIII->IV
nums[i] = 1;
nums[i - 1]++;
if(nums[i - 1] > 1){//两个以上的高位单位可以进位：VIV->IX
rstr[index - 1] = romans[i];
rstr[index++] = romans[i - 2];
}else{
rstr[index++] = romans[i];
rstr[index++] = romans[i - 1];
}
}else{
for(j = 0;j < nums[i];j++){
rstr[index++] = romans[i];
}
}
}
return rstr;
}
int main(){
int n = 3999;
char *s = intToRoman(n);
printf("%s\n",s);
free(s);
return 0;
}
时间： 2024-10-18 03:56:06