这两天参加了hihocoder上的小竞赛，下面把自己做的记录一下！（最痛心的是，开始竟然把main函数，写成了mian，浪费了将近一个小时时间，伤不起啊）

Description

Consider a string set that each of them consists of {0, 1} only. All strings in the set have the same number of 0s and 1s. Write a program to find and output the K-th string according to the dictionary order. If such a string doesn’t exist, or the input is not valid, please output “Impossible”. For example, if we have two ‘0’s and two ‘1’s, we will have a set with 6 different strings, {0011, 0101, 0110, 1001, 1010, 1100}, and the 4th string is 1001.

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10000), the number of test cases, followed by the input data for each test case.

Each test case is 3 integers separated by blank space: N, M(2 <= N + M <= 33 and N , M >= 0), K(1 <= K <= 1000000000). N stands for the number of ‘0’s, M stands for the number of ‘1’s, and K stands for the K-th of string in the set that needs to be printed as output.

Output

For each case, print exactly one line. If the string exists, please print it, otherwise print “Impossible”.

 

样例输入

32 2 22 2 74 7 47

样例输出

0101Impossible01010111011

这里给出我的答案

思路： 计算由N个0和M个1组成的数的最小值和最大值，然后判断从最小值开始，计算每个数包含一个的个数，到第K的时候，再将其转为包含N个0,和M个1的字符串，输出。（这是暂时想出的方法，以后会仔细思考一下，如果你有不同的方法，欢迎交流）

#include
     
      #include
      
       #include
       
        #include
        
         using namespace std;int fac(int n){    if(n == 1)        return 1;    else        return n*fac(n-1);}int numOne(int value){    int res = 0;    while(value)    {        if(value % 2 == 1)            res++;        value = value /2;    }    return res;}string outObj(int value,int len){    string res = "";    while(value)    {        if(value % 2 == 1)            res.append("1");        else            res.append("0");        value = value / 2;    }    reverse(res.begin(),res.end());    string app="";    for(int i=0; i < len - res.size(); i++)        app.append("0");    app.append(res);    return app;}int main(){    int T;    cin>>T;    int M,N,K;    vector
         
           result(T);    int minValue,maxValue;    for(int i=0; i
          
           >N>>M>>K; int len = M + N; minValue=0; maxValue=0; if(fac(N+M)/(fac(N) * fac(M)) < K) { cout<<"Impossible"<