poj 2282 The Counting Problem 按位统计

//poj 2282
//sep9
#include<iostream>
using namespace std;
typedef __int64 ll;
ll b[16];

ll f(ll n,ll x)
{
	ll left,m,ans=0;
	for(ll i=1;i<13;++i){
		left=n/b[i]-(x==0);
		ans+=left*b[i-1];
		m=(n%b[i]-n%b[i-1])/b[i-1];
		if(m>x)
			ans+=b[i-1];
		else if(m==x)
			ans+=n%b[i-1]+1; 
		if(b[i]>n)
			break;
	}	
	return ans;
}

int main()
{
	ll m,n;
	b[0]=1;
	for(int i=1;i<=12;++i)
		b[i]=b[i-1]*10;
	while(scanf("%I64d%I64d",&m,&n)&&(m+n)){
		if(m>n)
			swap(n,m); 
		for(ll i=0;i<10;++i)	
			printf("%I64d ",f(n,i)-f(m-1,i));		
		printf("\n");
	}
	return 0;
}