UC知道将大于0小于1000的阿拉伯数字转换为罗马数字
来源：百度知道 编辑：UC知道 时间：2024/06/02 19:53:54
这是网上的答案,但结果有点不正确,出现CX,谁能帮我把它改改?!
#include<stdio.h>
int main()
{
static char *a[][10]={"","I","II","III","IV","V","VI","VII","VIII","IX"
"","X","XX","XXX","XL","L","LX","LXX","LXXX","XCC",
"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"
}; /*建立对照表*/
int n,t,i,m;
printf("Please enter number:");
scanf("%d",&n); /*输入整数*/
printf("%d=",n);
for(m=0,i=1000;m<3;m++,i/=10)
{
t=(n%i)/(i/10); /*从高位向低位依次取各位的数字*/
printf("%s",a[2-m][t]); /*通过对照表翻译输出*/
}
printf("\n");
}
*运行结果
1. Please enter number:863
863=DCCCLXIII
2. Please ent
#include<stdio.h>
int main()
{
static char *a[][10]={"","I","II","III","IV","V","VI","VII","VIII","IX"
"","X","XX","XXX","XL","L","LX","LXX","LXXX","XCC",
"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"
}; /*建立对照表*/
int n,t,i,m;
printf("Please enter number:");
scanf("%d",&n); /*输入整数*/
printf("%d=",n);
for(m=0,i=1000;m<3;m++,i/=10)
{
t=(n%i)/(i/10); /*从高位向低位依次取各位的数字*/
printf("%s",a[2-m][t]); /*通过对照表翻译输出*/
}
printf("\n");
}
*运行结果
1. Please enter number:863
863=DCCCLXIII
2. Please ent
#include "Stdio.h"
#include "Conio.h"
void trans(int num,char *b[],int a[])
{int last=0;
int m;
int i;
while(num)
{m=num/a[last];
if(m<=0) last++;
else
{for(i=1;i<=m;i++)
printf("%s",*(b+last));
num%=a[last];
}
}
}
int main(void)
{ char *b[]={"m","cm","d","cd","c","xc","l","xl","x","ix","v","iv","i"};
int a[]={1000,900,500,400,100,90,50,40,10,9,5,4,1};
int num;
int m;
scanf("%d",&num);
trans(num,b,a);
getch();
return 0;
}