TOJ 3867: Cow IDs -- 递归

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本文详细阐述了如何确定指定位数且包含特定数量'1'的二进制数序列中第N个数的形式。通过数学分析和编程实现，解决了Farmer John在管理其奶牛编号时遇到的困惑。

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3867: Cow IDs

Time Limit(Common/Java):1000MS/3000MS Memory Limit:65536KByte
Total Submit: 41 Accepted:13

Description

Being a secret computer geek, Farmer John labels all of his cows with binary numbers. However, he is a bit superstitious, and only labels cows with binary numbers that have exactly K "1" bits (1 <= K <= 10). The leading bit of each label is always a "1" bit, of course. FJ assigns labels in increasing numeric order, starting from the smallest possible valid label -- a K-bit number consisting of all "1" bits. Unfortunately, he loses track of his labeling and needs your help: please determine the Nth label he should assign (1 <= N <= 10^7).

Input

* Line 1: Two space-separated integers, N and K.

Output

*Line 1: One number in binary form.

Sample Input

7 3

Sample Output

Hint

INPUT DETAILS:

Among all binary numbers containing exactly 3 "1" bits, FJ wants to output the 7th in increasing sorted order.

Source

USACO Feb. 2012

分析：用s[i][j]表示用i + 1个1和不超过j个0最多产生多少个数，那么lower_bound(s[k - 1], s[k - 1] + len, n)就是第n个数含有的0的个数，再算出剩下部分（从第2个1开始的部分）有多少个0，它们的差就是前两个1之间0的个数，然后递归剩下以1开头的部分。

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                          > rpqi; const int MAX_K = 10; const int MAX_N = 10000; const int MAX = 10000000; int s[MAX_K][MAX_N]; //s[i][j]: (i + 1) 1s and (<= j) 0s int cnt[MAX_K]; void ini(int n) { int i, j; for (i = 0; i < MAX_K; ++i) { s[i][0] = 1; } for (i = 0; i < MAX_N; ++i) { s[0][i] = i + 1; } for (i = 1; i < MAX_K; ++i) { for (j = 1; j < MAX_N; ++j) { s[i][j] = s[i][j - 1] + s[i - 1][j]; if (s[i][j] >= n) { cnt[i] = j + 1; break; } } } } void solve(int n, int k) { cout << 1; if (k == 1) { for (int i = 1; i < n; ++i) { cout << 0; } return; } else if (n == 1) { for (int i = 1; i < n; ++i) { cout << 1; } return; } int t1 = lower_bound(s[k - 1], s[k - 1] + cnt[k - 1], n) - s[k - 1]; n -= s[k - 1][t1 - 1]; int t2 = lower_bound(s[k - 2], s[k - 2] + cnt[k - 1], n) - s[k - 2]; for (int i = 0; i < t1 - t2; ++i) { cout << 0; } solve(n, k - 1); } int main(int argc, char *argv[]) { int n, k; cin >> n >> k; ini(n); solve(n, k); cout << endl; return 0; }

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