BZOJ 1799 同类分布(数位DP)

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-9
# define MOD 1024523
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=55;
//Code begin...

LL dp[20][165][165], p[20];
int wei[20];

LL dfs(int pos, int mod, int limit, int x, int P){
    if (x<0) return 0;
    if (pos==0) return mod==0&&x==0;
    if (!limit&&~dp[pos][mod][x]) return dp[pos][mod][x];
    int up=limit?wei[pos]:9;
    LL res=0;
    FOR(i,0,up) res+=dfs(pos-1,(mod+P-(i*p[pos-1]%P))%P,limit&&i==wei[pos],x-i,P);
    if (!limit) dp[pos][mod][x]=res;
    return res;
}
LL sol(LL x){
    int pos=0;
    while (x) wei[++pos]=x%10, x/=10;
    LL res=0;
    int top=min(162,pos*9);
    FOR(i,1,top) mem(dp,-1), res+=dfs(pos,0,1,i,i);
    return res;
}
int main ()
{
    LL a, b;
    p[0]=1; FO(i,1,20) p[i]=p[i-1]*10;
    scanf("%lld%lld",&a,&b);
    printf("%lld
",sol(b)-sol(a-1));
    return 0;
}