题目：​​https://www.lydsy.com/JudgeOnline/problem.php?id=3598​​

数位DP...东看西看 ​

好巧妙的思路啊！这样统计的东西就变得很简单了；

好美的 dfs！数位DP用 dfs 好像能变得很清楚。

代码如下：



#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
ll l,r,f[65][6005],ans;
int n,a[65],K;
ll dfs1(int pos,int s,bool lim)
{
    if(pos==0)return s;
    if(!lim&&f[pos][s]!=-1)return f[pos][s];
    int end=K-1; ll ret=0;
    if(lim)end=a[pos];
    for(int i=0;i<=end;i++)
        ret+=dfs1(pos-1,s+i*(pos-1),lim&&(i==end));
    if(!lim)f[pos][s]=ret;//!lim!
    return ret;
}
ll dfs(int pos,int s,int m,bool lim)
{
    if(s<0)return 0;//!
    if(pos==0)return s;
    if(!lim&&f[pos][s]!=-1)return f[pos][s];
    int end=K-1; ll ret=0;
    if(lim)end=a[pos];
    for(int i=0;i<=end;i++)
    {
        if(pos>=m)ret+=dfs(pos-1,s+i,m,lim&&(i==end));
        else ret+=dfs(pos-1,s-i,m,lim&&(i==end));
    }
    if(!lim)f[pos][s]=ret;
    return ret;
}
ll calc(ll x)
{
    int n=0;
    while(x)a[++n]=x%K,x/=K;
    memset(f,-1,sizeof f);
    ll ret=dfs1(n,0,1);
    for(int i=2;i<=n;i++)
    {
        memset(f,-1,sizeof f);
        ret-=dfs(n,0,i,1);
    }    
    return ret;
}
int main()
{
    scanf("%lld%lld%d",&l,&r,&K);
    printf("%lld\n",calc(r)-calc(l-1));
    return 0;
}