wust 1419 1419: We Love 01（ 计数问题）

1419: We Love 01

Time Limit: 1 Sec  Memory Limit: 128 MB   64bit IO Format: %lld
Submitted: 30  Accepted: 7
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Description

You have N ‘1’ and M ‘0’. You can use all of them to construct a binary number. For example you have 2 ‘1’ and 2 ‘0’. You can get 6 numbers “0110” “1100” “1010” “0011” “0101” “1001”.

Now, the problem is what’s the K-th smallest number?

Input

First line contains a number T, represent the number of test case.

Each case contains 3 numbers N M and K.

Output

For each case, output the K-th smallest number.

Sample Input

32 2 33 2 41 2 4

Sample Output

0110

01110

Impossible

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8//typedef __int64 ll;#define fre(i,a,b)  for(i = a; i <b; i++)#define free(i,b,a) for(i = b; i >= a;i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define ssf(n)      scanf("%s", n)#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define bug         pf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 55int c[N][N];void inint(){int i,j;fre(i,0,N)  fre(j,0,i+1)    if(j==0||i==j) c[i][j]=1;    else c[i][j]=c[i-1][j-1]+c[i-1][j];}void dfs(int n,int m,int k){     int i,j;     if(n==0) { fre(i,0,m) pf("1"); return ; }     if(m==0) { fre(i,0,n) pf("0"); return ; }     int temp=c[n+m-1][m];     if(temp>=k) { pf("0"); dfs(n-1,m,k); }     else { pf("1"); dfs(n,m-1,k-temp); }}int main(){int i,j,n,m,k,t;inint();sf(t);while(t--){sfff(m,n,k);if(c[n+m][m]<k){pf("Impossible\n");continue;}        dfs(n,m,k);        pf("\n");}   pf("\n");}/*32 2 33 2 41 2 4*/