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 18void 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");