微软2014在线笔试第二题 （对于问题中的k是有限制的，取值范围是1~9）

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 s​uch 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”. 

ps：这里面要求的K的取值是[1,1000000000],博主的程序限定了K的取值范围，改为[1,9]。

#include <stdio.h>
#include <string.h>
#include <math.h>
#define MAXWORDS 30000
#define NUM 18

void string_process(int s[], int start);
int countones(int tar_num);

int main()
{

char test_number[6];
char c;
int i, j, k, num ,tmp;
int test_cases[MAXWORDS];
int test[3];

memset(test_number, '*', sizeof(test_number));

//******************************************************输入数据****************************************//
printf("INPUT THE TEST CASES NUMBER YOU WANT  PS:it must ranged from 1 to 10000\n");
i =0;
while((c = getchar()) != '\n')
{
test_number[i] = c;
i++;
}

i = 0;
while(test_number[i] != '*')
{
i++;
}

num = 0;
tmp = i;
for ( j = 0; j < tmp; j++)
{
num = num + (test_number[j] - '0') * (int)pow(10, (int)(--tmp));
}

printf("the test cases number you want is %d\n", num);

k = 0;

for (i = 0; i < num; i++)
{
printf("INPUT THE %d th test case:\n", i+1);
j = 0;
while((c = getchar()) != '\n')
{
test[j] = c - '0';
j++;
}

for (j = 0; j<3; j++)
{
test_cases[k] = test[j];
k++;
}
}

tmp = k;
//***************************************************处理数据***********************************//
for (i = 0; i < num;i++)
{
printf("the %d th answer is :\n", i);
string_process(test_cases, i*3);
}
return 0;
}

void string_process(int s[], int start)
{
int i, j;
int sum , min, max;
int num_0 = s[start];
int num_1 = s[start+1];
int num = num_0 + num_1;
int k;
int k_number;
int a[NUM];
int m = num_0;
int n = num_1;
sum = jiecheng(m ,n);

if(s[start+2]>sum)
{
printf("IMPOSSIBLR\n");
return;
}
max = 0;
j = num;
min = 0;
k = 0;
for (i = 0; i < num_1; i++)
{
max = max + (int)pow(2,--j);
min = min + (int)pow(2,i);
}

for(i = min; i <= max; i++)
{
if ((j = countones(i)) == num_1)
{
k++;
if (k == s[start + 2])
{
k_number = i;
}
}
}

i = 0;
while(1)
{
a[i] = k_number%2;
k_number = k_number/2;
if (k_number == 0)
{
break;
}
i++;
}
while ((num-1) > i)
{
printf("0");
num--;
}
for (j = i; j >=0; j--)
{
printf("%d", a[j]);
}
printf("\n");

}

int countones(int tar_num)
{
int count = 0;
while(tar_num != 0)
{
tar_num = tar_num & (tar_num -1);
count++;
}
return count;
}

int jiecheng(int m, int n)
{
int c, sum;
int i;
c = m + n;

sum = 1;
for(i = 0; i < m;i++)
{
sum = sum * c;
c--;
}

for (i = 0; i < m;i++)
{
sum = sum / m;
m--;
}

return sum;
}

重要函数——int conutones(int n)：该函数的作用是算出一个整数的二进制形式包含多少个1

其中，如下一段代码表示将十进制转化为二进制，并打印出来：

i = 0;
while(1)
{
a[i] = k_number%2;
k_number = k_number/2;
if (k_number == 0)
{
break;
}
i++;
}
while ((num-1) > i)
{
printf("0");
num--;
}
for (j = i; j >=0; j--)
{
printf("%d", a[j]);
}
printf("\n");