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
3
2 2 2
2 2 7
4 7 47
样例输出
0101
Impossible
01010111011
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<iostream>
#include<string>
#include<vector>
#include<math.h>
using namespace std; int fac(int n)
{
if(n == )
return ;
else
return n*fac(n-);
}
int numOne(int value)
{
int res = ;
while(value)
{
if(value % == )
res++;
value = value /;
}
return res;
}
string outObj(int value,int len)
{
string res = "";
while(value)
{
if(value % == )
res.append("");
else
res.append("");
value = value / ;
}
reverse(res.begin(),res.end());
string app="";
for(int i=; i < len - res.size(); i++)
app.append("");
app.append(res);
return app;
}
int main()
{
int T;
cin>>T;
int M,N,K;
vector<int> result(T);
int minValue,maxValue;
for(int i=; i<T; i++)
{
cin>>N>>M>>K;
int len = M + N;
minValue=;
maxValue=;
if(fac(N+M)/(fac(N) * fac(M)) < K)
{
cout<<"Impossible"<<endl;
continue;
}
for(int j=; j<M; j++)
{
minValue = minValue + pow(, j);
}
for(int j=N; j<M+N; j++)
{
maxValue = maxValue + pow(, j);
}
int count = ;
int temp = ;
int obj = ;
for(int j = minValue; j <= maxValue; j++)
{
temp = numOne(j);
if(temp == M)
count++;
if(count == K)
{
obj = j;
break;
}
}
string res = outObj(obj,len);
cout<<res<<endl;
}
return ;
}