Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

 

Problem D: Bit-wise Sequence

Dr. DoLots is an expert in number theory and he is currently studying prime numbers. In fact, some people say that he has some formula which can generate the prime numbers in increasing order. But, his formula needs to order the non-negative integers according to the increasing number of ones in their binary representation. That is, all non-negative integers with m ones in their binary representation should be ordered before all the non-negative integers with m+1 ones.

More formally, lets define the non-negative integer sequence B_n such that for every non-negative integers m and n where n<m, 
            1. either O(B_n)< O(B_m)
            2. or O(B_n) = O(B_m) and B_n< B_m.
where O(n) is the number of ones in the binary representation of n.

Dr. DoLots wants you to write a program which computes the value of B_n for arbitrary values of n, 0� n� 2147483647. For simplicity, we consider only non-negative integers less than 2147483648 to appear in the sequence. You might be wondering why the doctor cannot do it himself. He is very busy and he wants to spent his time on more important things (That is what he says, actually, he is a bit lazy, but you can help him, can't you ?).

Input

Input consists of several lines specifying one non-negative integer value for n.

Output

Output should follow the same format as the input except that n is replaced by B_n.

Sample Input

0
1
2
31
32

Sample Output

0
1
2
1073741824
3

Arun Kishore

---

題目描述：

將 31 bits 的所有數字，按照 bits 個數大小再按照數字大小排序，得到它們的大小關係。

現在逆向反求，給定第幾個，輸出該數字為何。

#include <stdio.h>
long long C[50][50] = {};
void init() {
    C[0][0] = 1;
    int i, j, k;
    for(i = 1; i < 50; i++) {
        C[i][0] = 1;
        for(j = 1; j <= i; j++)
            C[i][j] = C[i-1][j] + C[i-1][j-1];
    }
}
int main() {
    init();
    long long n;
    while(scanf("%lld", &n) == 1) {
        long long ret = 0;
        int i, j, k;
        int ones;
        for(i = 0; i <= 31; i++) {
            if(n < C[31][i]) {
                ones = i;
                break;
            }
            n -= C[31][i];
        }
        for(i = 31; i >= 0; i--) {
            if(C[i][ones] > n)
                ret = ret << 1;
            else
                ret = ret << 1 | 1, n -= C[i][ones], ones--;
        }
        printf("%lld\n", ret);
    }
    return 0;
}