URAL 1209. 1, 10, 100, 1000... （规律 + 二分）

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

inputoutput

431476

0 0 1 0

题意：给定n个数ki，问1101001000......的第ki个位置的数字是0还是1.

解析：利用规律可推出第i个‘1’的位置编号为 i * (i - 1) / 2 + 1，利用二分判断询问的位置是否为1即可。

AC代码：

#include <cstdio>int check(long long x){    long long r = 1LL<<31, l = 0, mid, y;    while(l <= r){                      //二分判断是否为1        mid = l + (r - l)/2;        y = mid*(mid-1)/2 + 1;        if(y == x) return 1;        else if(y < x) l = mid + 1;        else r = mid - 1;    }    return l <= r;}int main(){    #ifdef sxk        freopen("in.txt", "r", stdin);    #endif //sxk    int n;    long long c;    scanf("%d", &n);    for(int i=0; i<n; i++){        scanf("%lld", &c);        printf("%d%c", check(c), i < n ? ' ' : '\n');    }    return 0;}