10166. 「一本通 5.3 练习 1」数字游戏

#include<algorithm>
#include<bitset>
#include<complex>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<iterator>
#include<limits>
#include<list>
#include<locale>
#include<map>
#include<memory>
#include<new>
#include<numeric>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<typeinfo>
#include<utility>
#include<valarray>
#include<vector>
#include<cctype>
#include<cerrno>
#include<cfloat>
#include<ciso646>
#include<climits>
#include<clocale>
#include<cmath>
#include<csetjmp>
#include<csignal>
#include<cstdarg>
#include<cstddef>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
using namespace std;
inline int read(){
    int res=0,f=1;char ch=getchar();
    while(ch<'0'ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') res=res*10+ch-'0',ch=getchar();
    return res*f;
}
inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x<10) putchar(x+'0');
    else{
        write(x/10);
        putchar(x%10+'0');
    }
}
//queue<int> q;
//set<int> s;
//priority_queue<int> q1;
//priority_queue<int,vector<int>,greater<int> > q2;
//list<int> l;
//stack<int> s;
int a,b,N,Dim[25],f[555][110];
int dfs(int x,int sum,int bj){//分别表示第几位，和为多少，上一位是否已经大于
    if(x==0){
        if(sum%N==0) return 1;
        else return 0;
    }
    if(bj==0&&f[x][sum]!=-1) return f[x][sum];//记忆化优化
    int res=0,ret=0;
    if(bj==1) res=Dim[x];
    else res=9;//k的范围
    for(int i=0;i<=res;i++)
        ret+=dfs(x-1,(sum+i)%N,bj&&i==res);//DP转移
    if(bj==1) return ret;
    else return f[x][sum]=ret;
}
int Get(int x){
    Dim[0]=0;
    while(x){
        Dim[++Dim[0]]=x%10;
        x/=10;
    }//把该数字拆开
    return dfs(Dim[0],0,1);//数位dp记忆化
}
void reset(){
    for(int i=0;i<=200;i++){
        for(int j=1;j<=110;j++) f[i][j]=-1;//因为每一次的询问的n不同，所以清空
    }
}
int main(){
    while(cin>>a){
        reset();
        b=read();N=read();
        write(Get(b)-Get(a-1));putchar('\n');//数位dp的基本框架
    }
    return 0;
}