#include <stdio.h>

void main() {
int c,n;
double num1,result;
char option;
/*printf("The number of conversions to be performed");*/
scanf("%d",&n);

for(c=1;c<=n;c++){
       /*printf("%d",c);*/

scanf(" %c",&option);
scanf(" %lf",&num1);

switch (option) {

case '1':result=num1*0.3048;
        printf("\n%.2lf feet equivalent to %.2lf meters\n",num1,result);
             break;
case '2':result=num1*1.852;
        printf("%.2lf nautical miles equivalent to %.2lf kilometers\n",num1,result);
             break;
case '3':result=num1*0.016387;
        printf("%.2lf cubic equivalent to %.2lf liters\n",num1,result);
             break;
case '4':result=num1*4.54609;
        printf("%.2lf gallons equivalent to %.2lf liters\n",num1,result);
             break;
case '5':result=num1*4046.8564;
        printf("%.2lf acers equivalent to %.2lf square meters\n",num1,result);
             break;
case '6':result=num1*0.4535924;
        printf("%.2lf pounds equivalent to %.2lf kilograms\n",num1,result);
             break;
case '7':result=num1*1016.0469;
        printf("%.2lf tons equivalent to %.2lf kilograms\n",num1,result);
             break;
case '8':result=num1*745.7;
        printf("%.2lf horsepower equivalent to %.2lf watts",num1,result);
             break;
case '9':result=num1*1.01387;
        printf("%.2lf horsepower equivalent to %.2lf metric horsepower",num1,result);
             break;
  }
}

}
请问如何限制n的次数为5？
也就是最多能转换5次。谢谢

/*printf("The number of conversions to be performed");*/
scanf("%d",&n);
在这里可以判断一下输入的n的值，如果n的值大于5，可以将该值变为5，也可以提示输入数字过大，请重新输入。
比如：
if(n>5)
n=5;
或者：
if(n>5)
printf("The number is larger than 5,please input one number again.然后用一个循环使其重新输入一个数，并在此判断。

[ 本帖最后由 m456m654 于 2009-10-19 12:36 编辑 ]

static int n=5;

#define n 5

直接设成5 不成么