acm的一道问题，不知道为什么Time limit exceeded

Question

Time Limit: 1.0 second
Memory Limit: 16 MB

Let's consider an infinite sequence of digits constructed of ascending powers of 10 written one after another. Here is the beginning of the sequence: 110100100010000… You are to find out what digit is located at the definite position of the sequence.
Input
There is the only positive integer number N in the first line, N < 65536. The i-th of N left lines contains the positive integer Ki — the number of position in the sequence. It's given that Ki < 231.
Output
You are to output N digits 0 or 1 separated with a space. More precisely, the i-th digit of output is to be equal to the Ki-th digit of described above sequence.
Sample

input
output

4
3
14
7
6
0 0 1 0

程序结果正确，格式无误，但就是不能通过，求教高手
#include<stdio.h>
#include<stdlib.h>
unsigned long x[65540];
int main()
{
long l;
unsigned long n,i,j,k;
scanf("%ld",&n);
k=0;
for (i=0;i<n;i++)
scanf("%ld",&x[k++]);
for (l=0;l<n;l++)
{
j=1;
for (i=0;i<x[l]-1;)
{
i+=j;
j++;
}
if(i==x[l]-1)
printf("1");
else printf("0");
if(l<n-1)printf(" ");
else printf("\n");
}
return 0;
}

Answer 1

错误时说你时间超时，就是程序运行的有点慢，我没看题意，但看了你的程序，感觉在i从1开始加，然后加这个循环可以省略的，你可以(x[l]-1-i)*2，然后开根号结果为y，然后判断y*(y+1)/2+i判断是否等于x[i]-1, 因为从1开始加的等差序列，求和不就是首项加末项，乘以项数，除以2。你可以把题目链接发一下。。。 我觉得ki是小于2的31次方吧，不然会超时，所以数值，用long long比较好，乘以2之后会不会数据溢出。。。下面代码应该能AC吧
#include <queue>
#include <stack>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <limits.h>
#include <string.h>
#include <string>
#include <algorithm>

using namespace std;

int main()
{
int ncases,n;

scanf("%d",&ncases);

while( ncases-- )
{
scanf("%d",&n);
int k = (int)(sqrt((double)2*(n-1)));
if( k*(k+1ll)/2+1 == n )
printf("1");
else
printf("0");
if( ncases ) printf(" ");
}

return 0;

}