Exemplo n.º 1
0
void print_permutation(int n, int h, int pos)
{
	if( pos==n )
	{
		for( int i = 0 ; i!=n ; ++i )
		{
			printf("%d",out[i]);
		}
		printf("\n");	
	}
	else
	{
		int usedzero=0,usedone=0;
		int numone=h,numzero=n-h;
		for( int i = 0 ; i!=pos ; ++i )
		{
			if( out[i]==1 )
			{
				usedone++;
			}
			else
				usedzero++;
		}
		if( usedzero < numzero )
		{
			out[pos]=0;
			print_permutation(n, h, pos+1);
		}
		if( usedone < numone )
		{
			out[pos]=1;
			print_permutation(n,h,pos+1);
		}
	}
}	
Exemplo n.º 2
0
Arquivo: c104.c Projeto: gsrr/Programs
void DFS(int index)/*index->x used->y*/
{
        if(index == 9)
        {
                int tmp_max = print_permutation();
                if(tmp_max > max)
                {
                        max = tmp_max;
                }
                return;
        }

        int a;
        for(a = 1; a <= 8 ; a++) /*a->y*/
        {
                if(isValid(index, a))
                {
                        board[index][0] = index;
                        board[index][1] = a;
                        used[a] = 1;
                        DFS (index+1);
                        used[a] = 0;
                }
        }
}
Exemplo n.º 3
0
void print_permutation(int n, int length, char* permutation)
{
	if( n==length )
	{
		permutation[n]='\0';
		printf("%s\n",permutation);
	}
	else
	for( int i = 0 ; i!=length ; ++i )
	{
		if( i==0 || input[i]!=input[i-1] )
		{
			int used=0,num=0;
			for( int j = 0 ; j!=length ; ++j )
			{
				if( input[j]==input[i] )
				{
					num++;
				}
			}
			for( int j = 0 ; j!=n ; ++j )
			{
				if( permutation[j]==input[i] )
				{
					used++;
				}
			}
			if( used<num )
			{
				permutation[n]=input[i];
				print_permutation(n+1, length, permutation);
			}
		}
	}
}
Exemplo n.º 4
0
void print_permutation(char *wordRa){
	static count;
	char singleWord[2];
	char aheadWord[10],currentwordRa[10];
	int index,wordRalength,i;
	wordRalength=strlen(wordRa);
	if(wordRalength>2){
		
		strcpy(currentwordRa,wordRa);
		for(i=0;i<wordRalength;i++){
			singleWord[0]=currentwordRa[0];
			singleWord[1]='\0';
			strcat(toBePrinted,singleWord);
			strcpy(aheadWord,currentwordRa+1);
			print_permutation(aheadWord);
			index=strlen(toBePrinted)-1;
			toBePrinted[index]='\0';
			rotateLeft(currentwordRa);
		}
		
	}else
	{
		//printf("\n%d)TOBEpTelse=%s",++count,toBePrinted);
		printf("\n%s",toBePrinted);
		printf("%s",wordRa);
		//printf("\n%d)TOBEpTelse=%s",++count,toBePrinted);
		printf("\n%s",toBePrinted);
		printf("%c%c",wordRa[1],wordRa[0]);
	}
}
Exemplo n.º 5
0
int main(){
  memset(vis, 0, sizeof(vis));
  int n, A[MAXN];
  scanf("%d",&n);
  print_permutation(n, A, 0);
  printf("n=%d,way2 time used: %.2lf\n",n, (double)clock()/CLOCKS_PER_SEC);
  return 0;
}
Exemplo n.º 6
0
main(){
	char word[20];
	scanf("%s",word);
	//gets(word);
	//rotateLeft(word);
	//printf("%s",word);
	print_permutation(word);
}
Exemplo n.º 7
0
int main(){
	int n; 
	scanf("%d",&n);
	for(int i = 0  ;i <  n;i ++)
	{
		scanf("%d",&b[i]);
	}
	print_permutation(n,b,a,0);
	return 0 ;
}
Exemplo n.º 8
0
void print_permutation(int n, int A[], int cur){ // cur == current
  if(cur == n) {
    for(int i=0; i<n; i++) printf("%d ",A[i]);
    putchar('\n');
  }
  else for(int i=1; i<=n; i++) if(!vis[i]){
      vis[i] = 1;
      A[cur] = i;
      print_permutation(n, A, cur+1);
      vis[i] = 0;
  }
}
Exemplo n.º 9
0
int main(int argc, char *argv[])
{
	int n,h,num;
	scanf("%d",&num);
	while( num-- )
	{
		scanf("%d %d",&n,&h);
		print_permutation(n,h,0);
		if(num)
		printf("\n");
	}
}
Exemplo n.º 10
0
int main(int argc, char *argv[])
{
	int n;
	scanf("%d\n",&n);
	while( n-- )
	{
		fgets(input, 20, stdin);
		int length = strlen(input)-1;
		std::sort(input, input+length);
		print_permutation(0, length, permutation);
		printf("\n");	
	}
}
Exemplo n.º 11
0
int main(){
  freopen("722.in", "r", stdin);
  freopen("722time.txt", "a", stdout);
  int n, A[MAXN], P[MAXN];
  scanf("%d",&n);
  for(int i = 0; i < n; i++) scanf("%d",&P[i]);
  //n = 4;
  std::sort(P,P+n);
  //for(int i= 0;i< n;i++) printf("%d ",P[i]);
  print_permutation(n, P, A, 0);
  printf("n=%d, way1 time used: %.4lfs\n",n, (double)clock()/CLOCKS_PER_SEC);
  return 0;
}
Exemplo n.º 12
0
// 输出1~n的全排列
void print_permutation(int n, int* A, int cur) {
  int i, j;
  if(cur == n) { // 递归边界
    for(i = 0; i < n; i++) printf("%d ", A[i]);
    printf("\n");
  } else for(i = 1; i <= n; i++) { // 尝试在A[cur]中填各种整数i
    int ok = 1;
    for(j = 0; j < cur; j++)
      if(A[j] == i) ok = 0; // 如果i已经在A[0]~A[cur-1]出现过,则不能再选
    if(ok) {
      A[cur] = i;
      print_permutation(n, A, cur+1); // 递归调用
    }
  }
}
Exemplo n.º 13
0
void print_permutation(int n, int P[], int A[], int cur){
  if(cur == n) {
    for(int i=0; i<n; i++) ;//printf("%d ",A[i]);
    //putchar('\n');
  }
  else for(int i=0; i<n; i++) if(!i || P[i] != P[i-1]) {
    //if的意�? 对于重复元素(排序后已经挨在一起),只试它们中的第一个,其余跳过
    //P[0]是必然要试的:若有多个等于P[0]的值,P[0]必为第一�?0-1=-1不在数组下表范围�?;若只有一个,更理所当然地要试P[0]�?!i即为i==0)
    int c1 = 0, c2 = 0;
    for(int j=0; j<cur; j++) if(A[j] == P[i]) c1++;
    for(int j=0; j<n; j++) if(P[j] == P[i]) c2++;
    if(c1 < c2) {
      A[cur] = P[i];
      print_permutation(n, P, A, cur+1);
    }
  }
}
Exemplo n.º 14
0
int main(int argc, char* argv[])
{
  // input
  uint64_t r; // desired rank
  uint32_t n; // length of set
  uint32_t k; // length of permutation
  
  uint32_t* p; // identity permutation
  uint32_t* unranked; // unranked permutation
  
  if (argc < 3 || argc > 4)
  {
    printf("Usage: %s r n <k>\n", argv[0]);
    return 1;
  }
  
  r = strtoull(argv[1], (char**)NULL, 10);
  n = strtoul(argv[2], (char**)NULL, 10);
  
  if (argc == 4)
  {
    k = strtoul(argv[3], (char**)NULL, 10);
  }
  else
  {
    k = n;
  }
  
  if (k > n)
  {
    printf("Error: value of k must be less than or equal to n.\n");
    return 1;
  }
  
  p = create_identity_permutation(n);
  unranked = unrank(n, k, r, p);
    
  print_permutation(k, unranked);
    
  free(p);
  
  return 0;
}
Exemplo n.º 15
0
void print_permutation(int *A,int cur,int n){
    if(cur==n){
        for(int i=0;i<n;i++){
            printf("%d ",A[i]);
        }
        printf("\n");
    }else{
        for(int i=1;i<=n;i++){
            bool ok = true;
            for(int j=0;j<cur;j++){
                if(A[j]==i){
                    ok = false;
                    break;
                }
            }
            if(ok){
                A[cur] = i;
                print_permutation(A,cur+1,n);
            }
        }   
    }
}
Exemplo n.º 16
0
//freopen("/home/plac/problem/input.txt","r",stdin);
//freopen("/home/plac/problem/output.txt","w",stdout);
void print_permutation(int n,int *P,int *A,int cur)
{
	int i , j; 
	if(cur == n)
	{
		for(int i = 0 ;i < n ;i ++) printf("%d ",A[i]);
		printf("\n");
	}else	
		for(i = 0;i < n;i ++)
		if(!i || P[i] != P[i-1])
		{
			int c1 = 0 ,c2 = 0;
			for(j = 0 ;  j < cur ; j ++) if(A[j] == P[i]) c1 ++ ;
			for(j = 0 ;j < n ; j ++) if(P[j] == P[i]) c2++;
			if(c1 < c2)
			{
				A[cur] = P[i];
				print_permutation(n,P,A,cur+1);
			}

		}

}
Exemplo n.º 17
0
int main() {
    int n;
    scanf("%d",&n);
    print_permutation(A,0,n);
    return 0;
}
Exemplo n.º 18
0
int main() {
  print_permutation(4, A, 0); 
  return 0;
}