FZU 2113（数位dp）

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdlib>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define LL long long
#define mod 10007
#define inf 0x3f3f3f3f
#define N 100010
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
LL dp[22][22];
int dig[22];
//pos表示第pos位，sum表示前面已包含1的个数，limit表示后面是否可以任意填
LL dfs(int pos,int sum,bool limit)
{
    if(pos==0)return sum;
    if(!limit&&~dp[pos][sum])return dp[pos][sum];
    int len=limit?dig[pos]:9;
    LL ans=0;
    for(int i=0;i<=len;i++)
    {
        if(i==1)ans+=dfs(pos-1,sum+1,i==len&&limit);
        else ans+=dfs(pos-1,sum,i==len&&limit);
    }
    if(!limit)dp[pos][sum]=ans;
    return ans;
}
LL solve(LL x)
{
    int len=0;
    while(x)
    {
        dig[++len]=x%10;
        x/=10;
    }
    LL ans=dfs(len,0,1);
    return ans;
}
int main()
{
    LL a,b;
    while(cin>>a>>b)
    {
        memset(dp,-1,sizeof(dp));
        printf("%I64d\n",solve(b)-solve(a-1));
    }
}