1.c

编程求N<=100时

#include "stdio.h"
#include "conio.h"

main()

{
    long a[256],b,d;
    int n,i,j,m,k;

  l1:  m=1;
      a[0]=1;
    printf("Please input a number:");
    scanf("%d",&n);
    for(i=2;i<=n;i++)
    { d=0 ;
      for(j=0;j<m;j++)
      { b=a[j]*i+d;
        a[j]=b % 1000000;
        d=b/1000000;
      }


       /*当累乘结果的最后一个元素与要累乘的数据有进位时就让m加1，
       可避免不必要的乘0运算*/
       if(d!=0)
       { a[j]=d;
         m=m+1;
        }
     }
      /*以下完成计算结果的输出*/
     printf("%d!=",n);
     printf("%ld\t",a[m-1]);
     for(i=m-2;i>=0;i--)
     {
      if(a[i]>99999)
      {
       printf("%ld\t",a[i]);
       continue;
       }
       if(a[i]>9999)
      {
       printf("0"); 
       printf("%d\t",a[i]);
       continue;
       }

      if(a[i]>999)
      {  printf("00");
       printf("%ld\t",a[i]);

       continue;
       }
       if(a[i]>99)
      {  printf("000");
       printf("%ld\t",a[i]);


       continue;
       }

       if(a[i]>9){
       printf("0000");
       printf("%ld\t",a[i]);
       continue;
       }
       printf("00000");
       printf("%ld\t",a[i]);



     }
     printf("\nPress '1'to continue  or '0' to exit!\n");
     scanf("%d",&k);
     if(k==1)
     goto l1;
    else
      exit(0);



    getch();


}