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<<31)-1;
    LL down=1;
    LL mid;
    LL m;
    while(up>=down){
        mid=(up+down)/2;
        m=mid*(mid-1)/2+1;
        if(m==k) return 1;
        else if(m>k) up=mid-1;
        else down=mid+1;
    }
    return 0;
}


int main()
{
    int n,k;
    //freopen("data.txt","r",stdin);
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&k);
        if(i) printf(" ");
        printf("%d",check(k));
    }
    printf("\n");
    return 0;
}