c将数字转换为罗马数字的程序

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function是printf和scanf。 我已经写了代码，但我不知道如何打印出诸如“25,13,​​93,66”等数字。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  3000) printf("%d can not be represented in Roman numerals. \n", number); return 0; } 

Answer 1

您似乎尝试使用以下百=数字+ 0连接数字;

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，查看代码会更容易，因为这样可以更详细地概述代码流以及范围落在何处。

我希望这些小技巧可以帮助:)。

Answer 2

 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); } 

Answer 3

 #include  #include  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 

Answer 4

 #include  #include  #include  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; } 

Answer 5

 #include 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"); 

}

很长（43行），但简单吧!!

Answer 6

这个问题确实令人生畏，至少在初学者层面……，因为你提到只使用基本function，那么使用我的解决方案可以轻松打印1000以下的数字：

 # include 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=(tr)/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++; } }