codeforces 55d//Beautiful numbers// Codeforces Beta Round #51

標籤：int   mes   swa   case   sse   typedef   queue   個數   bool   

題意：一個數能整除它所有的位上的數字（除了0），統計這樣數的個數。

注意離散化，為了速度更快需存入數組尋找。

不要每次memset，記錄下已有的長度下合格個數。

數位dp肯定是從高位到低位。

記錄數字已經有多大，還有lcm，遞迴傳下去。

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <iomanip>#include <cstring>#include <map>#include <queue>#include <set>#include <cassert>using namespace std;const double EPS=1e-8;const int SZ=5050,INF=0x7FFFFFFF;typedef long long lon;lon dp[20][3000][50],index[3000];//2520vector<lon> ls;lon gcd(lon x,lon y){    if(x<y)swap(x,y);    for(;;)    {        lon rem=x%y;        if(rem==0)return y;        x=y;        y=rem;    }}lon lcm(lon x,lon y){    return x*y/gcd(x,y);}void getid(){    for(lon i=0;i<ls.size();++i)    {        index[ls[i]]=i;    }}void lsh(){    for(lon i=1;i<pow(2,10);++i)    {        lon cur=1;        for(lon j=0;j<9;++j)        {            if(i&(1<<j))cur=lcm(cur,j+1);        }        ls.push_back(cur);    }    sort(ls.begin(),ls.end());    ls.erase(unique(ls.begin(),ls.end()),ls.end());    getid();}lon dfs(lon pos,lon cur,lon prelcm,lon limit,string &str){    if(pos==str.size())return cur%prelcm==0;    if(!limit&&dp[str.size()-pos-1][cur][index[prelcm]]!=-1)return dp[str.size()-pos-1][cur][index[prelcm]];    lon res=0;    lon up=limit?str[pos]-‘0‘:9;    for(lon i=0;i<=up;++i)    {        lon curlcm=prelcm;        if(i)curlcm=lcm(i,curlcm);        res+=dfs(pos+1,(cur*10+i)%2520,curlcm,limit&&str[pos]-‘0‘==i,str);    }    if(!limit)dp[str.size()-pos-1][cur][index[prelcm]]=res;    return res;}lon work(string &str){    return dfs(0,0,1,1,str);}bool chk(string &str){    lon num=0,curlcm=1;    for(lon i=0;i<str.size();++i)    {        num=(num*10+str[i]-‘0‘)%2520;        if(str[i]!=‘0‘)curlcm=lcm(curlcm,str[i]-‘0‘);    }    return num%curlcm==0;}int main(){    std::ios::sync_with_stdio(0);    //freopen("d:\\1.txt","r",stdin);    lon casenum;    cin>>casenum;    lsh();    memset(dp,-1,sizeof(dp));    for(lon time=1;time<=casenum;++time)    {        string ll,rr;        cin>>ll>>rr;        lon res=work(rr)-work(ll);        if(chk(ll))++res;        cout<<res<<endl;    }    return 0;}

 

codeforces 55d//Beautiful numbers// Codeforces Beta Round #51