Round A APAC Test 2016 Problem A. Googol String

Problem A. Googol String

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Small input 7 points	Solve A-small
Large input 12 points	Solve A-large

Problem

A "0/1 string" is a string in which every character is either 0 or 1. There are two operations that can be performed on a 0/1 string:

• switch: Every 0 becomes 1 and every 1 becomes 0. For example, "100" becomes "011".
• reverse: The string is reversed. For example, "100" becomes "001".

Consider this infinite sequence of 0/1 strings:

S₀ = ""

S₁ = "0"

S₂ = "001"

S₃ = "0010011"

S₄ = "001001100011011"

...

S_N = S_N-1 + "0" + switch(reverse(S_N-1)).

You need to figure out the Kth character of S_googol, where googol = 10¹⁰⁰.

Input

The first line of the input gives the number of test cases, T. Each of the next T lines contains a number K.

Output

For each test case, output one line containing "Case #x: y", where x is the test case number (starting from 1) and y is the Kth character of S_googol.

Limits

1 ≤ T ≤ 100.

Small dataset

1 ≤ K ≤ 10⁵.

Large dataset

1 ≤ K ≤ 10¹⁸.

Sample

Input	Output
4 1 2 3 10

按题中要求构造字符串，问第k个是0还是1,

其实，我们可以发现，每个字符串都是成倍+1的增长，如果是在中心点（也就是2^n -1）则一定就是0，否则的话，按中心点对称，就是转化为2^n -2 - k;这样，很快就会变的很小或者为中心点，再记录翻转的次数，也就是这样转化的次数，如果是奇数次就是1，否则就是0，直接得到答案了。复杂度为o(log(k));

#define N 205
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,T;
ll p[100];
ll k;
int main()
{
    //freopen("A-large.in", "r", stdin);
    //freopen("A-large.out", "w", stdout);
    p[0] = 1;
    For(i,1,100){
        p[i] = p[i-1] * 2;
        printf("%lld %d\n",p[i],i);
    }
     while(S(T)!=EOF)
    {
        For(ta,1,T+1){
            scanf("%lld",&k);k--;
            int flag = 0,num = 0;
            bool isEnd = false;
            while(k){
                //printf("%lld k ",k);
                For(i,0,100){
                    if(p[i] - 2 >= k){
                        flag = i;
                        break;
                    }
                }
                if(k == p[flag] - 2ll - k){
                    isEnd = true;
                }
                if(isEnd) break;
                num++;


                k = p[flag] - 2ll - k;
            }
            //printf(" num %d ",num);
            if(!(num & 1))
                printf("Case #%d: 0\n",ta);
            else
                printf("Case #%d: 1\n",ta);
        }

    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}