f(n)=10f(n-1) + 10^(n-1)  n > 1
f(n)=1                    n = 1
即f(n)=n10^(n-1)

    0 --  9999 这里得减去一些多余的0
10000 -- 10999
11000 -- 11999
12000 -- 12099
12100 -- 12199
12200 -- 12299
12300 -- 12309
12310 -- 12319
12320 -- 12329
12330 -- 12339
12340 -- 12345

#include<fstream>
using namespace std;

unsigned long countArray[10] = {0};

void Count( unsigned long num = 0 )
{	
	if( num<0 ) return;	
	
	char number[11] = {0};	
	sprintf( number, "%d", num );
	
	int i;
	for (i=0; i<10; ++i)
		countArray[i] = 0;
	
	int cursor = 0;
	int numberCount = 0;	
	while( number[cursor] != '\0' )
	{
		// Add one digital at the end, the count of the number will be about 10 double. 
		// So the count of left digital will be approximately 10 times.		
		for(i=0; i<10; ++i )
			countArray[i] *= 10;
		
		// But not all			
		// For 2=>23, '1' 10 double, '2' only 3+1,
		// For 2=>29, '2' will 10 dobule too.
		for( i=0; i<cursor; ++i )
			countArray[ number[i]-'0' ] -= '9' - number[cursor];
		
		// There are some new digital for the last digit.
		// Add one for each digital.
		for( i=0; i<10; ++i )
			countArray[i] += numberCount;
		
		// Not all again.
		for(i=1; i<=( number[ cursor ] - '0' ); ++i)
			++countArray[i];
		
		// Get the count for previous part.
		numberCount *= 10;
		numberCount += number[ cursor ] - '0';
		
		++cursor;
	}
	
	++countArray[0];
}

void main(void)
{	
	ifstream fin("in.txt");
	ofstream fout("out.txt");
	
	int testCount;
	fin>>testCount;
	while(--testCount)
	{		
		unsigned long num;
		fin>>num;	
		Count(num);
		
		//Output
		fout<<countArray[0];
		for(int i=1; i<10; i++ )
			fout<<","<<countArray[i];
		fout<<endl;
	}	
}

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