腾讯举办的一场比赛——1007. Binary Numbers

题目描述
XDTICers all like Binary numbers. They want to know the Kth(K <= 10000000) binary numbers whose sum of all ‘1’ digits is N(N <= 10).
输入
There are several test cases
The first line, an integer T indicates the test cases.
Next T lines, two integers K, N as mentioned above.
输出
Output the Kth number in the binary number form.
样例输入
1
7 3
样例输出
10110

此题显然就是全排列的思想。
但是因为K值太大，不能一开始就全排列，否则会超时。
那么我们先剪枝。
找找规律发现，当K值和N值一定时，那么一共的可能种数是S=C(N-1)(N-1)+C(N-1)(N)+C(N-1)(N+1)+...。
所以当我们的S快接近K值(S要小于K)，那么再开始对那时候的情况排列组合。

以下是我AC的代码。

#include <set>
#include <map>
#include <list>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <vector>
#include <iomanip>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;

int pa[100];

int pl(int a, int b)
{
 int ss = 1;
 for (int i = 0; i < a; i++)
 {
  ss = ss * (b - a + i + 1) / (i + 1);
 }
 return ss;
}

int main()
{
 int t;
 
 scanf("%d", &t);
 while (t--)
 {
  int k, n;
  scanf("%d%d", &k, &n);
  if (n == 1)
  {
   printf("1");
   for (int i = 0; i < k - 1; i++)
   {
    printf("0");
   }
   printf("\n");
   continue;
  }
  int s = 1;
  int p = 0;
  while (1)
  {
   int t = pl(n - 1, n + p);
   if (s + t > k) break;
   s += t;
   p++;
  }
  p += n;
  for (int i = 0; i <= p - n; i++)
  {
   pa[i] = 0;
  }
  for (int i = p - n + 1; i < p; i++)
  {
   pa[i] = 1;
  }
  while (1)
  {
   s++;
   if (s == k)
   {
    printf("1");
    for (int i = 0; i < p; i++)
    {
     printf("%d", pa[i]);
    }
    printf("\n");
    break;
   }
   next_permutation(pa, pa + p);
  }
 }
    
    return 0;
}