Codeforces 55D Beautiful numbers 数位DP

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自从去年4月份学姐教完我数位DP，还是第一次写出这么漂亮的代码，也是醉了。

首先你要知道sum%(x*n) %x == sum%x，这样就可以在dfs的时候记录2到9的最小公倍数2540取余了。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-8)
#define LL long long
#define ULL unsigned long long
#define _LL __int64
#define INF 0x3f3f3f3f
#define Mod 6000007

//** I/O Accelerator Interface .. **/
#define g (c=getchar())
#define d isdigit(g)
#define p x=x*10+c-'0'
#define n x=x*10+'0'-c
#define pp l/=10,p
#define nn l/=10,n
template<class T> inline T& RD(T &x)
{
    char c;
    while(!d);
    x=c-'0';
    while(d)p;
    return x;
}
template<class T> inline T& RDD(T &x)
{
    char c;
    while(g,c!='-'&&!isdigit(c));
    if (c=='-')
    {
        x='0'-g;
        while(d)n;
    }
    else
    {
        x=c-'0';
        while(d)p;
    }
    return x;
}
inline double& RF(double &x)      //scanf("%lf", &x);
{
    char c;
    while(g,c!='-'&&c!='.'&&!isdigit(c));
    if(c=='-')if(g=='.')
        {
            x=0;
            double l=1;
            while(d)nn;
            x*=l;
        }
        else
        {
            x='0'-c;
            while(d)n;
            if(c=='.')
            {
                double l=1;
                while(d)nn;
                x*=l;
            }
        }
    else if(c=='.')
    {
        x=0;
        double l=1;
        while(d)pp;
        x*=l;
    }
    else
    {
        x=c-'0';
        while(d)p;
        if(c=='.')
        {
            double l=1;
            while(d)pp;
            x*=l;
        }
    }
    return x;
}
#undef nn
#undef pp
#undef n
#undef p
#undef d
#undef g

using namespace std;

LL dp[22][257][2521];

int num[21];

LL dfs(int up,int site,int sta,int mod,int bot)
{
    if(site > bot)
    {
        for(int i = 2;i <= 9; ++i)
            if((sta&(1<<(i-2))) && mod%i != 0)
                return dp[site][sta][mod] = 0;
        return dp[site][sta][mod] = 1;
    }

    if(up == 0 && dp[site][sta][mod] != -1)
        return dp[site][sta][mod];

    int i;

    LL ans = 0;

    if(up == 0)
    {
        ans += dfs(0,site+1,sta,(mod*10+0)%2520,bot);
        ans += dfs(0,site+1,sta,(mod*10+1)%2520,bot);
        for(i = 2;i <= 9; ++i)
            ans += dfs(0,site+1,sta|(1<<(i-2)),(mod*10+i)%2520,bot);
        return dp[site][sta][mod] = ans;
    }
    else
    {
        if(0 <= num[site])
            ans += dfs(0 == num[site] ? 1 : 0,site+1,sta,(mod*10+0)%2520,bot);
        if(1 <= num[site])
            ans += dfs(1 == num[site] ? 1 : 0,site+1,sta,(mod*10+1)%2520,bot);
        for(i = 2;i <= num[site]; ++i)
            ans += dfs(i == num[site] ? 1 : 0,site+1,sta|(1<<(i-2)),(mod*10+i)%2520,bot);

        return ans;
    }
}

LL Cal(LL x)
{
    if(x == 0)
        return 1;

    int i;

    for(i = 20;x; --i)
        num[i] = x%10,x /= 10;

    return dfs(1,i+1,0,0,20);
}

int main()
{

    int T,icase;

    int n,i,j;

    scanf("%d",&T);

    memset(dp,-1,sizeof(dp));

    for(icase = 1;icase <= T; icase++)
    {
        LL l,r;

        cin>>l>>r;

        cout<<Cal(r)-Cal(l-1)<<endl;
    }

    return 0;
}

Codeforces 55D Beautiful numbers 数位DP

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原文地址：http://blog.csdn.net/zmx354/article/details/42880403