1001WA

时间关系只实现了其中一部分的功能  现在的程序可以实现一个大数的平方

 #include <stdio.h>
 #include <string.h>
 void mypower(int num2);
 int findflag();
 int myflag = ;
 char num1[];
 int num2;
 char a[] = {};
 char b[] = {};
 int c[] = {};
 int d[] = {};
 int main(void)
 {
     scanf("%s%d",num1,&num2);
     myflag = findflag();
     mypower(num2);
     return ;
 }

 void mypower(int num2)
 {
     int i,j,k,s;
     for(s = ;s < num2;s++)
     {
         memset(c,,sizeof(c));
         if(b[] == '\0')
         {
             for(i = ;a[i] != '\0';i++)
             {
                 b[i] = a[i];
             }
         }
         for(i = ;a[i] != '\0';i++)
             for(j =;b[j] != '\0';j++)
             {
                 c[i+j] += (a[i] - '') * (b[j] - '');

             }
         for(k = i+j-;k > ;k--)
         {
             d[k] = (c[k-] % );
             c[k-] += c[k-] /;
         }
         d[] = (c[] / );
         d[] = (c[] % );
         if(d[] != )
         {
             for(k = ;k < i+j;k++)
                 b[k] = d[k];
         }else
         {
             for(k = ;k < i+j;k++)
                 b[k-] = d[k];
         }

     }
     myflag *= num2;
     if(d[] != )
     {
         for(s =;s < myflag;s++)
         {
             b[i+j-s] = b[i+j--s];
             b[i+j--s] = '.';
         }
         for(i = ;i<k+;i++)
         {
             if(b[i] == '.')
             {
                 printf("%c",b[i]);
                 continue;
             }
             printf("%d",b[i]);
         }
     }else
     {
         for(s =;s < myflag;s++)
         {
             b[i+j--s] = b[i+j--s];
             b[i+j--s] = '.';
         }
         for(i = ;i<k;i++)
         {
             if(b[i] == '.')
             {
                 printf("%c",b[i]);
                 continue;
             }
             printf("%d",b[i]);
         }
     }

 }

 int findflag()
 {
     int i,j,flag = ;
     for(i = ;num1[i] != '\0';i++)
     {
         if(num1[i] == '.')
         {
             flag = i;
             continue;
         }
         a[j++] = num1[i];
     }
     flag = j - flag;
     return flag;
 }