[codeforces1073E]Segment Sum

time limit per test : 1 second
memory limit per test : 256 megabytes

分数：2300
You are given two integers $l$ and $r (l≤r)$. Your task is to calculate the sum of numbers from $l$ to $r$ (including $l$ and $r$) such that each number contains at most k different digits, and print this sum modulo $998244353$.

For example, if $k=1$ then you have to calculate all numbers from l to r such that each number is formed using only one digit. For $l=10$, $r=50$ the answer is $11+22+33+44=110$.

Input

The only line of the input contains three integers $l, r$ and $k (1≤l≤r<10^{18},1≤k≤10)$ — the borders of the segment and the maximum number of different digits.

Output

Print one integer — the sum of numbers from $l$ to $r$ such that each number contains at most $k$ different digits, modulo $998244353$

.
Examples
Input

10 50 2

Output

1230

Input

1 2345 10

Output

2750685

Input

101 154 2

Output

2189

Note

For the first example the answer is just the sum of numbers from $l$ to $r$ which equals to $50⋅51/2−9⋅10/2=1230$. This example also explained in the problem statement but for $k=1$.

For the second example the answer is just the sum of numbers from $l$ to $r$ which equals to $2345⋅2346/2=2750685$.

For the third example the answer is $101+110+111+112+113+114+115+116+117+118+119+121+122+131+133+141+144+151=2189$.

题意：
给定l,r,k，询问[l,r]中所有x的总和,x是满足：组成x的所有数码的种类数不超过k。

题解：
数位dp
设dp(x)为 $l=1,r=x$的答案
$f[i][S][j][l]$表示第 $i$位, $j$表示生成的数字串是否小于 $x$, $l$表示是否大于 $0$的数字个数。
$d[i][S][j][l]$表示第 $i$位, $j$表示生成的数字串是否小于 $x$, $l$表示是否大于 $0$的数字之和。
然后就可以数位dp交替更新了。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll MOD=998244353LL;
int Cuming(int S){
    int c=0;
    for(int i=0;i<10;i++){
        if(S&1)c++;
        S>>=1;
    }
    return c;
}
ll l,r;
int k;
int len,sp[24];
ll f[24][1<<11][2][2],g[24][1<<11][2][2];
ll calc(ll x){
    memset(sp,0,sizeof(sp));
    memset(f,0,sizeof(f));
    memset(g,0,sizeof(g));
    ll ret=0;
    len=0;
    f[1][0][0][0]=1;
    if(x==0)sp[++len]=0;
    while(x){
        sp[++len]=x%10;
        x/=10;
    }
    reverse(sp+1,sp+len+1);
    for(int i=1;i<=len;i++){
        for(int j=0;j<2;j++){
            for(int l=0;l<2;l++){
                for(int S=0;S<(1<<10);S++){
                    int Lim;
                    if(j)Lim=9;
                    else Lim=sp[i];
                    if(!f[i][S][j][l])continue;
                    for(int now=0;now<=Lim;now++){
                        int nS;
                        if(l||now)nS=(S|(1<<now));
                        else nS=S;
                        f[i+1][nS][j||(now<Lim)][l||now]+=f[i][S][j][l];
                        f[i+1][nS][j||(now<Lim)][l||now]%=MOD;
                        g[i+1][nS][j||(now<Lim)][l||now]+=(((g[i][S][j][l]*10)%MOD)+(now*f[i][S][j][l])%MOD)%MOD;
                        g[i+1][nS][j||(now<Lim)][l||now]%=MOD;
                    }
                }
            }
        }
    }
    for(int i=0;i<2;i++){
        for(int j=0;j<2;j++){
            for(int l=0;l<(1<<10);l++){
                if(Cuming(l)<=k){
                    ret+=g[len+1][l][i][j];
                    ret%=MOD;
                }
            }
        }
    }
    return ret;
}
int main(){
    scanf("%lld%lld%d",&l,&r,&k);
    ll ansl=calc(l-1);
    ll ansr=calc(r);
    printf("%lld\n",(ansr-ansl+MOD)%MOD);
    return 0;
}