Timus - 1209 - 1, 10, 100, 1000...

先上题目：

1209. 1, 10, 100, 1000...

Time limit: 1.0 second
Memory limit: 64 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 integer N in the first line (1 ≤ N ≤ 65535). The i-th of N left lines contains the integer Ki — the number of position in the sequence (1 ≤ Ki ≤ 231 − 1).

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

　　题意：按照1,10,100,1000```的顺序将数字排在一起，从左往右，(10^i)<=(2^31-1)，问第k位是0还是1。

　　有人可以推出公式直接求第k位是0还是1，但是这里我用的方法不一样，首先我们可以得出第x个1出现的位置会是哪里:(x-1)*x/2+1=k，如果k符合这个条件就说明第k位是一个1否则就是0。不过如果枚举x来找k的话一定会超时，所以我们可以二分查找把x找出来，如果可以找出x说明是1，否则就是0了。

上代码：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #define LL long long
 using namespace std;

 int check(LL k){
     LL up=(1LL<<)-;
     LL down=;
     LL mid;
     LL m;
     while(up>=down){
         mid=(up+down)/;
         m=mid*(mid-)/+;
         if(m==k) return ;
         else if(m>k) up=mid-;
         else down=mid+;
     }
     return ;
 }

 int main()
 {
     int n,k;
     //freopen("data.txt","r",stdin);
     scanf("%d",&n);
     for(int i=;i<n;i++){
         scanf("%d",&k);
         if(i) printf(" ");
         printf("%d",check(k));
     }
     printf("\n");
     return ;
 }

1209