请看下我的算法是否有问题谢谢
#include<stdio.h>
void main()
{
  int i;
  int j;
  int a[2000]={0};
  int num=0;
  for(i=0;i<2000;i++)
  {
      scanf("%d",&a[i]);
      while(a[i]!=0)
      {
          int flag=0;
          for(j=1;;j++)
          {
              int m=0,sum=0;
              if(a[i]==(j*(j+1))/2)
              {
                  flag=1;
                  break;
              }
              if(a[i]>(j*(j+1))/2)
              {
                  break;
              }
          }
          if(flag)
          {
              int m=0,sum=0;
              int k;
              printf("%d is a triangular number.\n",a[i]);
              for(k=1;;k++)
              {
                  sum+=k;
                  if(sum<a[i])
                     m++;
                  if(sum==a[i])
                     break;
              }
              for(k=0;k<=m;k++)
              {
                  printf("*");
              }
              printf("\n");
              break;

          }
          else
              printf("%d is not a triangular number.\n",a[i]);
          break;
          
      }
      break;
  }  
}

题目是这样的
A triangle can be made up out of dots, like the ones shown below:



The number of dots in each one is called a triangular number. More precisely, a number is said to be triangular if it is of the form &frac12; n(n+1). The numbers 1, 3, 6, 10, and 15 are all triangular.

Given a number, determine if it is triangular. If it is, draw the corresponding right triangle, if it is not, say so.

Input
A positive integer less than 2000. Each case will be on a separate line. A zero (0) denotes the end of input.

Output
The phrase “<num> is a triangular number.” followed by a right triangle of “*” characters, or the phrase “<num> is not a triangular number.” The triangle should be oriented with the hypotenuse on the right and the right angle in the lower left.

Sample Input
3
4
6
0
Sample Output
3 is a triangular number.
*
**
4 is not a triangular number.
6 is a triangular number.
*
**
***

#include <stdio.h>
#include <math.h>
int main(void)
{
    int n;
    while (scanf("%d", &n) == 1)
    {
        int r = sqrt(n = 8 * n + 1);
        if ((r * r == n) && (r & 1))
            printf("Yes,%d\n", (r - 1) / 2);
        else printf("No\n");
    }
    return 0;
}

有没有问题你在OJ上submit一次不就知道了

能不能讲解一下？谢谢

看懂了题目肯定会做。这题目数学方法显然可以快很多，我是O(1)，你是O(n)