版权声明：写了自己看的，看不懂不能怪我emmmm。 https://blog.csdn.net/qq_40858062/article/details/83997523
题目链接：http://codeforces.com/problemset/problem/1073/E
思路：数位dp按位求贡献算和
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(998244353)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
ll a[20],n,m,k,b[20];
Pll dp[20][1028];
Pll dfs(int pos,int qw,bool zero,bool limit)
{
if(!pos) return Pll(1,0);
if(!limit&&dp[pos][qw].second!=-1) return dp[pos][qw];
int up=limit?a[pos]:9;
Pll res=Pll(0,0),tmp;
FOR(i,0,up)
{
int zz=qw|((zero||i)<<i);
if(__builtin_popcount(zz)>k) continue;
tmp=dfs(pos-1,zz,zero||i,limit&&i==a[pos]);
res.first=(res.first+tmp.first)%mod;
res.second=(res.second+tmp.second+1ll*i*b[pos-1]%mod*tmp.first%mod)%mod;//tmp.first表示后面有多少个状态满足条件
}
if(!limit) dp[pos][qw]=res;
return res;
}
ll query(ll n)
{
a[0]=0;
while(n) a[++a[0]]=n%10,n/=10;
return dfs(a[0],0,0,1).second;
}
int main()
{
cin.tie(0);
cout.tie(0);
REW(dp,-1);
b[0]=1;FOR(i,1,19) b[i]=b[i-1]*10ll%mod;
sl(n),sl(m),sl(k);
ll x=query(m);
ll y=query(n-1);
printf("%I64d\n",((x-y)%mod+mod)%mod);
return 0;
}