2025年1th 整数罗马数转换

科技前沿 • 2025-04-03 12:26 • 阅读 53
大家好，我是讯享网，很高兴认识大家。
代码很简单，采用查表法做。所以直接贴出来吧。
#include &lt;stdio.h&gt; #include &lt;stdlib.h&gt; #include “string.h” char lut_1[10][8] = {“I”,“II”,“III”,“IV”,“V”,“VI”,“VII”,“VIII”,“IX”}; char lut_10[10][8] = {“X”,“XX”,“XXX”,“XL”,“L”,“LX”,“LXX”,“LXXX”,“XC”}; char lut_100[10][8] = {“C”,“CC”,“CCC”,“CD”,“D”,“DC”,“DCC”,“DCCC”,“CM”}; char lut_1000[10][8] = {“M”,“MM”,“MMM”}; /* * * 讯享网func_def: to separate int numbers into bits , store them to a int buffer return length example : when num_int is 1234 , num_bit_buf[0] &#61; 1 , num_bit_buf[1] &#61; 2 , num_bit_buf[2] &#61; 3 , num_bit_buf[3] &#61; 4 , return 4  
  
    
   
   * */ int separate(int num_int , int * num_bit_buf) { int i = 0,j = 0; if(num_int &gt; 999) { num_bit_buf[i] = num_int / 1000; //thousand num_bit_buf[i+1] = (num_int % 1000 ) /100; //hundrand num_bit_buf[i+2] = (num_int % 100 ) / 10; num_bit_buf[i+3] = (num_int % 10) ; return 4; } else if(num_int &gt; 99) { num_bit_buf[i] = num_int / 100; num_bit_buf[i+1] = (num_int % 100) / 10; num_bit_buf[i+2] = (num_int % 10); return 3; } else if(num_int &gt; 9) { num_bit_buf[i] = num_int / 10 ; num_bit_buf[i+1] = num_int % 10; return 2; } else { num_bit_buf[i] = num_int; return 1; } } 
  
 int int_2_roma(int num_int , char * roma_buf) { int num_bit_buf[4] , len &#61; 0 , i&#61;0; len &#61; separate(num_int,num_bit_buf); //printf(&#34;lens is %d\n&#34;, len); if(len&lt;1) { printf(&#34;error when separate!\n&#34;); return -1; } 
 #if 0 //set to 0 when we stop debug
讯享网 讯享网for(i&#61;0;i&lt;len;i&#43;&#43;) { printf(&#34;num_bit_buf[%d] is %d\n&#34;, i, num_bit_buf[i]); } 
 #endif switch(len) { case 4: { strcpy(roma_buf,lut_1000[num_bit_buf[0]-1]); strcat(roma_buf,lut_100[num_bit_buf[1]-1]); strcat(roma_buf,lut_10[num_bit_buf[2]-1]); strcat(roma_buf,lut_1[num_bit_buf[3]-1]); break; } case 3: { strcpy(roma_buf,lut_100[num_bit_buf[0]-1]); strcat(roma_buf,lut_10[num_bit_buf[1]-1]); strcat(roma_buf,lut_1[num_bit_buf[2]-1]); break; } case 2: { strcpy(roma_buf,lut_10[num_bit_buf[0]-1]); strcat(roma_buf,lut_1[num_bit_buf[1]-1]); break; } case 1: { strcpy(roma_buf,lut_1[num_bit_buf[0]-1]); //printf(&#34;src %s\n&#34;, lut_1[num_bit_buf[0]]); //puts(roma_buf); break; } } return 0; 
 } int main(int argc, char argv) { 讯享网char *p; char roma_buf[20]; int number_input&#61;strtol(argv[1], &amp;p, 10); if(argc!&#61;2 || number_input &gt; 3999){ printf(&#34;Usage: int2roma number(0-3999)\n&#34;); return -1; } //printf(&#34;number_input is %d\n&#34;, number_input); int_2_roma(number_input,roma_buf); puts(roma_buf); return 1; 
 }
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2025年1th 整数 罗马数 转换

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2025年1th 整数罗马数转换
