C实现topcoder的SquareDigits(编译通过)

Problem Statement
 ***Note: Please keep programs under 7000 characters in length. Thank you

Class Name: SquareDigits
Method Name: smallestResult
Parameters: int
Returns: int

Define the function S(x) as the sum of the squares of the digits of x.
For example: S(3)=3*3=9 and S(230)=2*2+3*3+0*0=13.

Define the set T(x) to be the set of unique numbers that are produced by
repeatedly applying S to x. That is: S(x), S(S(x)), S(S(S(x))), etc...
For example, repeatedly applying S to 37:
S(37)=3*3+7*7=58.
S(58)=5*5+8*8=89.
S(89)=145.
S(145)=42.
S(42)=20.
S(20)=4.
S(4)=16.
S(16)=37.
Note this sequence will repeat so we can stop calculating now and:
T(37)={58,89,145,42,20,4,16,37}.
However, note T(x) may not necessarily contain x.

Implement a class SquareDigits, which contains a method smallestResult. The
method takes an int, n, as a parameter and returns the smallest int, x, such
that T(x) contains n.

The method signature is (be sure your method is public):
int smallestResult(int n);

TopCoder will ensure n is non-negative and is between 0 and 199 inclusive.

Examples:
If n=0: S(0) = 0, so T(0)={0}, so the method should return 0.

If n=2: T(0) through T(10) do not contain the value 2. If x=11, however:
S(11)=1*1+1*1=2, so T(11) contains 2, and the method should return 11.

If n=10: T(0) through T(6) do not contain 10. If x=7:
S(7)=49.
S(49)=97.
S(97)=130.
S(130)=10.
S(10)=1.
and it starts to repeat...
so T(7) is {49,97,130,10,1}, which contains 10, and the method should return 7.

n=1 -> x=1
n=19 -> x=133
n=85 -> x=5
n=112 -> x=2666
Definition
 Class:  SquareDigits
Method:  smallestResult
Parameters:  int
Returns:  int
Method signature:  int smallestResult(int param0)
(be sure your method is public)

 
 //以下是C代码

#include <stdio.h>
#include <cstdlib>
#include <stdlib.h>
#include <string.h>
#define N 50

main()
{
 int num=0;
 int rem=0;
 int sum=0;
 int i=0,j=0;
 int t;
 int pnum[N]={0};
 
 printf("输入整数:\n");
 scanf("%d",&num);
 
 t=num;    
 while(num<0 || num>199)
   {
    printf("输入不合法。\n");
    system("pause");
    exit(1);   
   }
  
  loop1: 

 do
   {
    rem = num%10;
    sum += rem*rem;
    num =num/10;  
   }while(num!=0);
 
   //printf("sum=%3d  ",sum); system("pause");   //测试sum输出是否正确
   //printf("i=%d  ",i);
   num=sum;                                    //将sum赋值给num，num为新产生的数
  
   //printf("\n");
   //printf("pnum[%d]=%d  ",i,sum);             //测试存储的sum是否正确                                            
   //printf("sum0=%d  ",sum);system("pause");    //测试sum=0

   if(i==N)
    {
      printf("在50个范围内没有找到，退出");
      system("pause");
      exit(1);  
    }
 
  // 将sum和t（即输入的原始数字）比较是否相等 ，相等写入数字后停止计算
  while(t==sum)
    {
     pnum[i]=sum;
     for(j=0;j<=i;j++)
      {
       printf("%d  ",pnum[j]);  
      }
     printf("\n出现和原始数字相等，退出");
     system("pause"); 
     exit(1);     
    }

  //将sum和已经计算出写入到数组的数字做比较判断，判断是否重复，重复则停止写入到数组
  if(i==0)               //当产生第一个数时，直接存储
    {
    // printf("------");
     pnum[i]=sum;i++;
     sum=0;goto loop1; 
    }
  else
    {
     for(j=0;j<i;j++)
      {
        if(pnum[j]==sum)
          {
            //printf("++++++++\n");
            goto loop2;  
          }       
       }
      pnum[i]=sum;
      sum=0;  i++;
     
      //printf("========");
      goto loop1;
     }
  
 
  loop2:
   for(j=0;j<i;j++)
     {
      printf("%d  ",pnum[j]);  
     }
     
  system("pause");  
}