链接：https://ac.nowcoder.com/acm/contest/897/J
来源：牛客网

Binary Number
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
As a programmer, you are probably familiar with the binary representation of integers. That is, write an integer x as
∑
a
i
2
i
∑ai2i, where each
a
i
ai is either 0 or 1. Particularly, n-digit binary number can be written as
∑
n
−
1
i
=
0
a
i
2
i
∑i=0n−1ai2i, in which
a
n
−
1
an−1 must equal to 1.
This time, to test your mastery of binary numbers, Leg Han raises a problem to you.
Among all n-digit binary numbers whose amount of 1 is m, please print the k-th smallest one.
It is guaranteed that k is legal.
输入描述:
The first line contains an integer number T, the number of test cases.

i
t
h
ith of each next T lines contains three integers n, m, k(
1
≤
m
≤
n
≤
31
,
1
≤
k
≤
2
×
10
8
1≤m≤n≤31,1≤k≤2×108).
输出描述:
For each test case print the \(k\)-th smallest one.
示例1
输入
复制
2
5 2 2
5 3 3
输出
复制
10010
10110

题意：

思路：
我是反着来求的，要求第k小，我把所有情况sum算出来,令id=sum-k+1.然后求第id大的

从高位到低位，每一位我们考虑如果填了1，剩下还有种组合情况x，是否还够达到第id大，如果够就填1，不够就填0，并且id-=x。这样操作即可。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2) { ans = ans * a % MOD; } a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/


ll getpc(ll m, ll n) //m 中 选 n 个
{
    long long ans = 1;
    for (long long k = 1; k <= n; k++) {
        ans = (ans * (m - n + k)) / k;
    }
    return ans;
}
ll n, m, k;
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);

    int t;
    gbtb;
    cin >> t;
    while (t--) {
        cin >> n >> m >> k;
        ll sum = getpc(n - 1, m - 1);
        ll id = sum - k + 1ll;// 反过来求第id大的数。
        ll temp = 0ll;
        ll rm = m;// 剩余1的数量
        for (ll i = n; i > 0; --i) {
            ll x = getpc(i - 1, rm - 1);
            if (x >= id && rm) {
                cout << 1;
                rm--;
            } else {
                id -= x;
                cout << 0;
            }
            cout << flush;
        }
        cout << endl;
    }




    return 0;
}

inline void getInt(int *p)
{
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    } else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}