[USACO12FEB]牛的IDCow IDs

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.

输出格式：

输入输出样例

输入样例#1：

7 3 

输出样例#1：

10110 

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，可以二分求m

这题最坑的是为了防止溢出longlong，所以要将k分情况制定二分右界

如果做到第i位，还要填j个1，那么这一位填0的方案数就是C(i-1,j)，即还剩i-1位可以填j个1的方案数。

如果这个数小于n，那么这一位填1，并且n要减去这个数，否则这一位填0。

myys

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
ll n;
int k,cnt,m,ans[1001];
ll C(int x,int y)
{int i;
  if (x>y) return 0;
  ll s=1;
  for (i=y;i>y-x;i--)
    s*=i;
  for (i=x;i>=2;i--)
    s/=i;
  return s;
}
int main()
{int i;
  cin>>n>>k;
  if (k==1)
    {
      cout<<1;
      for (i=n-1;i>=1;i--)
    {
      printf("0");
    }
      return 0;
    }
  int l=1,r;
  if (k>=10)
  r=600;
  else if (k>=7)
    r=1000;
  else r=8000;
  while (l<=r)
    {
      int mid=(l+r)/2;
      if (C(k,mid)>=n) m=mid,r=mid-1;
      else l=mid+1;
    }
  cnt=0;
  for (i=m;i>=1;i--)
    {
      ll p=C(k,i-1);
      if (p<n) 
    {
      k--;
      ans[i]=1;
      n-=p;
      if (cnt==0)
        {
          cnt=i;
        }
    }
      if (k==0||n==0)
    {
      break;
    }
    }
  for (i=cnt;i>=1;i--)
    printf("%d",ans[i]);
}