微软2014年4月实习生招聘机试题 2.K-th string

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本文探讨了在特定数量的0和1排列中找到第K个字符串的方法，并提供了实现细节和示例输入输出。

《2.K-th string》

Time Limit: 10000ms
Case Time Limit: 1000ms
Memory Limit: 256MB

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”.

Sample In

3
2 2 2
2 2 7
4 7 47

Sample Out

0101
Impossible
01010111011

本来想用组合法求解，但可分析出对 n+m<=33且 m、n值不确定的情况下即使用long类型涉及到求组合数也会溢出，无论是用 C(m,n)=A(m,n) / n! ，还是用公式C(m,n)=m!/((m-n)! * n!)，事实上第二个公式更容易溢出，一般求较小数的组合时我们均会采用第一个公式来编写程序；

个人给出的一个解如下：

#include <iostream>
#include <fstream>
using namespace std;

int main(){
	long tmp=0;
	//tmp = combination(5,2);
	//printf("%ld",tmp);
	int testCaseCnt=0;
	int m,n,k;
	int mm,nn,kk;
	int index=0;
	int sumCase=0;
	char charArea[34];
	long recComb;//record combination 记录上次组合数
	fstream fstr("testCase.txt");
	fstr>>testCaseCnt;

	while(testCaseCnt--){
		memset(charArea,0,sizeof(charArea));
		fstr>>n>>m>>k;


		if(k==1){
			while(n--)
				cout<<'0';
			while(m--)
				cout<<'1';
			cout<<endl;
			continue;
		}
		index = 0;
		sumCase = 0;//
		mm=m;
		//nn=n;
		//kk=k;
		//设index =k时，组合数为 com,推到得 index =K+1时，对应组合数为com * (mm+k)/(k+1)，mm为当前剩余'1'个数
		recComb = 1; 
		while(mm){
			if(index == 0)
				recComb = 1;
			else 
				//recComb *=((mm+index-1)/(index));
				recComb = (recComb*(mm+index-1))/index;
			sumCase += recComb;
			if(k > sumCase){
				index++;
				if(index >n)
					break;
			}
			else if(k==sumCase){
				while(mm){
					charArea[index+mm-1]='1';
					mm--;
				}
				break;
			}
			else if(k < sumCase){
				charArea[index+mm-1]='1';
				sumCase -= recComb;
				recComb = 0;
				mm--;
				index = 0;
			}
		}
		if(sumCase < k){
			cout<<"Impossible"<<endl;
			continue;
		}
		for(int i=0;i<m+n;i++){
			if(charArea[i]!='1')
				charArea[i] = '0';
		}
		for(int i =0;i<(m+n)/2;i++){
			char tmp=charArea[i];
			charArea[i]=charArea[(m+n-1)-i];
			charArea[(m+n-1)-i]=tmp;

		}
		charArea[m+n]=0; //
		//charArea[m+n]=0;
		cout<<charArea<<endl;
	}
	fstr.close();
	
}