-
- 12通过
- 67提交
- 题目提供者lin_toto
- 标签USACO2012
- 难度普及/提高-
- 时空限制1s / 128MB
最新讨论更多讨论
- 谁能解释一下这个样例啊....
题目描述
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).
FJ给他的奶牛用二进制进行编号,每个编号恰好包含K 个"1" (1 <= K <= 10),且必须是1开头。FJ按升序编号,第一个编号是由K个"1"组成。
请问第N(1 <= N <= 10^7)个编号是什么。
输入输出格式
输入格式:- Line 1: Two space-separated integers, N and K.
输入输出样例
输入样例#1:
7 3
输出样例#1:
10110
分析:首先有一个很简单的结论:一个只有0和1的数字串,只有1对数字串大小有影响,0没有影响。很简单证明,大小取决于1的位置和数量。
这道题有一个限制:第一位必须是0,那么我们先将这个串用足够大小保存,足够大的话我们可以添加前导0,到最后从第一个非0位输出即可,也就是说我们要找到一个m,使得C(m,k) >= n,这个可以用二分实现,我们先弄一个m位的全是0的串。然后考虑C(i-1,k)的意义,即还剩i-1位可以填k个1的方案数,也就是说我们还有C(i,k)个不同大小的数,如果C(i-1,k) < n,则说明剩下的数还不够n个,我们不能找到第n大的数,于是我们在i位填1,那么这个数就是能够出现的C(i-1,k)个数中最大的,n-=C(i-1,k),k--,如果C(i-1,k) >= n,说明后面还能找到第n大的,我们填0即可,就这样模拟一下就好了。
不过这个组合数会非常大,还会爆long long,需要分类讨论进行二分.
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; long long n, k, f[50][20], m; long long num[110000], cnt; long long Combination(long long n, long long m) { long long ans = 1; for (long long i = n; i >= (n - m + 1); --i) ans *= i; while (m) ans /= m--; return ans; } int main() { scanf("%lld%lld", &n, &k); if (k == 1) { for (int i = n; i; i--) { if (i == n) printf("1"); else printf("0"); } return 0; } else { if (k == 10) { long long l = 1, r = 600; while (l <= r) { long long mid = (l + r) >> 1; if (Combination(mid, k) >= n) { m = mid; r = mid - 1; } else l = mid + 1; } } else { if (k >= 7) { long long l = 1, r = 1000; while (l <= r) { long long mid = (l + r) >> 1; if (Combination(mid, k) >= n) { m = mid; r = mid - 1; } else l = mid + 1; } } else { long long l = 1, r = 7000; while (l <= r) { long long mid = (l + r) >> 1; if (Combination(mid, k) >= n) { m = mid; r = mid - 1; } else l = mid + 1; } } } for (long long i = m; i; i--) { long long t = Combination(i - 1, k); if (t < n) { num[i] = 1; n -= t; k--; if (!cnt) cnt = i; } if (!k || !n) break; } for (long long i = cnt; i; i--) printf("%d", num[i]); } return 0; }