BZOJ 1799 同类分布(数位DP)

时间：2024-07-09 12:03:56

给出a,b，求出[a,b]中各位数字之和能整除原数的数的个数。1<=a<=b<=1e18.

注意到各位数字之和最大是153.考虑枚举这个东西。那么需要统计的是[0,a-1]和[0,b]内各位数字之和为x且能整除x的数字个数。

那么我们只需要数位dp一波即可。

令dp[pos][i][x]表示有pos位且数字之和为x的数mod P=i的数字个数。

则转移方程显然可得。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

LL dp[][][], p[];

int wei[];

LL dfs(int pos, int mod, int limit, int x, int P){

    if (x<) return ;

    if (pos==) return mod==&&x==;

    if (!limit&&~dp[pos][mod][x]) return dp[pos][mod][x];

    int up=limit?wei[pos]:;

    LL res=;

    FOR(i,,up) res+=dfs(pos-,(mod+P-(i*p[pos-]%P))%P,limit&&i==wei[pos],x-i,P);

    if (!limit) dp[pos][mod][x]=res;

    return res;

}

LL sol(LL x){

    int pos=;

    while (x) wei[++pos]=x%, x/=;

    LL res=;

    int top=min(,pos*);

    FOR(i,,top) mem(dp,-), res+=dfs(pos,,,i,i);

    return res;

}

int main ()

{

    LL a, b;

    p[]=; FO(i,,) p[i]=p[i-]*;

    scanf("%lld%lld",&a,&b);

    printf("%lld\n",sol(b)-sol(a-));

    return ;

}

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