UESTC 548 Cow IDs （二进制的排列组合数 STL）_548的2进制是多少-优快云博客

本文链接：https://blog.youkuaiyun.com/aozil_yang/article/details/51078151

本文介绍了一种算法，用于找到第N小的二进制数，该数由K个1和若干个0组成。通过计算排列组合并使用next_permutation函数来生成所需的二进制数。

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Cow IDs

Time Limit: 3000/1000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)

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≤107).

Input

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

Output

Line 1: Output the Nth label he should assign

Sample input and output

Sample Input	Sample Output
7 3	10110

Source

USACO Feb 2012

大体题意：

给你N 和K ，让你求一个二进制数，包含K个1，其余只能是0，求第N小的数！

思路：

先求出此时的N和K时，有多少个0 有多少1组成，N不断的减，（最后N的意义就是在当前0和当前1的数第几个小的）然后构建一个最小的数，比如说有3个1 和2 个0 组成吧，那么最小的数就是10011，然后利用next_permutation函数求n次即可！

最后，输出字符串即可！

注意：

数组要开大点，100W差不多就可以了！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1000000 + 10;
char s[maxn];
int main(){
	int n,k,m=0;
	scanf("%d%d",&n,&k);
	int star = 1;
	for ( ; n > star ; ++m){
		n-=star;
		star = star * (k+m)/(m+1);
	}
	s[0]='1';
	for (int i = 1; i < k + m; ++i){
		if (i <= m)s[i]='0';
		else s[i] = '1';
	}
	s[k+m]=0;
	while(--n && next_permutation(s+1,s+k+m));
	printf("%s\n",s);
	return 0;
}