PAT 1049 Counting Ones 数位dp

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
int bit[15],dp[15][15][2];
int nm,n;
int solve(int nn,int up,int bd)
{
	if(dp[nn][up][bd]!=-1)return dp[nn][up][bd];
	if(nn == 1)
	{
		if(up>=1)return dp[nn][up][bd] = 1;
		else return dp[nn][up][bd] = 0;
	}
	if(!bd)return dp[nn][up][bd] = (up+1)*solve(nn-1,up,bd)+(int)pow(10,nn-1);
	else 
	{
		dp[nn][up][bd] = up*solve(nn-1,9,0)+solve(nn-1,bit[nn-1],1);
		if(up == 1)dp[nn][up][bd]+=n%(int)pow(10,nn-1)+1;
		else if(up)dp[nn][up][bd]+=(int)pow(10,nn-1);
	    return dp[nn][up][bd];
	
	}
}
int main()
{
	int tn;
	while(scanf("%d",&n)!=EOF)
	{
	nm = 1;
	tn = n;
	memset(dp,-1,sizeof(dp));
	while(tn)
	{
		bit[nm++] = tn%10;
		tn/=10;
	}
	nm--;
	printf("%d\n",solve(nm,bit[nm],1));
    }
	return 0;
}