数位DP

AHOI2009 同类分布

先枚举各位数字之和作为模数再DP

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
#include<cmath>
#include<map>
#define l(x) (x<<1)
#define r(x) ((x<<1)+1)
#define IL inline
#define reg register
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
 
int i,len,mod;
LL a,b;
int z[20];
LL mi[19],f[20][200][200];
 
IL int Abs(int x){return (x<0)?-x:x;}
IL void Swap(int &a,int &b){a^=b^=a^=b;}
IL int Min(int a,int b){return (a<b)?a:b;}
IL int Max(int a,int b){return (a>b)?a:b;}
 
IL LL read(){
    LL p=0,f=1; char    c=getchar();
    while (c<48||c>57)  {if (c=='-')    f=-1;   c=getchar();}
    while (c>=48&&c<=57)    p=(p<<1)+(p<<3)+c-48,c=getchar();
    return p*f;
}
 
inline LL Mi(LL x,LL y,int MOD){
    LL p=x,t=1,Res=1;
    for (;t<=y;(t&y)?Res=(Res*p)%MOD:0,p=(p*p)%MOD,t<<=1);
    return Res;
}
 
IL LL Dfs(int pos,int lead,int limit,LL tot,LL Sum){
    reg LL i=0,up=0,sum=0,res=0;
    if (pos>len){
        if (Sum!=mod) return 0;
        return (!(tot%mod))?1:0;
    }
    res=(tot*(mi[len-pos+1]%mod))%mod;
    if (f[pos][res][Sum]!=-1 && !limit && !lead)    return f[pos][res][Sum];
    up=(limit)?z[len-pos+1]:9;
    for (i=0;i<=up;i++){
        if (!i && lead) sum+=Dfs(pos+1,1,limit&&(i==up),tot,Sum);
        else if (i && lead) sum+=Dfs(pos+1,0,limit&&(i==up),(tot<<3)+(tot<<1)+i,Sum+i);
        else sum+=Dfs(pos+1,0,limit&&(i==up),(tot<<3)+(tot<<1)+i,Sum+i);
    }
    return (!limit && !lead)?f[pos][res][Sum]=sum:sum;
}
 
IL LL Cal(LL x){
    reg LL i=0,ans=0;
    len=0;
    while (x){z[++len]=x%10;    x/=10;}
    for (i=1;i<=9*len;i++){
        memset(f,-1,sizeof(f));
        mod=i;
        ans+=Dfs(1,1,1,0,0);
    }
    return ans;
}
 
int main(){
    #ifdef __Marvolo
    freopen("zht.in","r",stdin);
    freopen("zht.out","w",stdout);
    #endif
    a=read();   b=read();
    for (mi[1]=10,i=2;i<=18;i++)    mi[i]=mi[i-1]*10;
    cout<<Cal(b)-Cal(a-1)<<endl;
    return 0;
}

花神的数论题

和上面那个差不多,枚举1的数量

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
#include<cmath>
#include<map>
#define l(x) (x<<1)
#define r(x) ((x<<1)+1)
#define IL inline
#define reg register
#define LL long long
#define MOD 10000007
#define INF 0x3f3f3f3f
using namespace std;
 
LL n,len,l;
LL z[75],f[75][75];
 
IL int Abs(int x){return (x<0)?-x:x;}
IL void Swap(int &a,int &b){a^=b^=a^=b;}
IL int Min(int a,int b){return (a<b)?a:b;}
IL int Max(int a,int b){return (a>b)?a:b;}
 
IL int read(){
    int p=0,f=1;    char    c=getchar();
    while (c<48||c>57)  {if (c=='-')    f=-1;   c=getchar();}
    while (c>=48&&c<=57)    p=(p<<1)+(p<<3)+c-48,c=getchar();
    return p*f;
}
 
IL LL Mi(LL x,LL y){
    LL p=x,t=1,Res=1;
    for (;t<=y;(t&y)?Res=(Res*p)%MOD:0,p=(p*p)%MOD,t<<=1);
    return Res;
}
 
IL LL Dfs(int pos,int pre,int lead,int limit,int s){
    LL sum=0,up=0,i=0;
    if (pos>len)    return (s==l);
    if (!lead && !limit && f[pos][s]!=-1)   return f[pos][s];
    up=(limit)?z[len-pos+1]:1;
    for (i=0;i<=up;i++){
        if (!i && lead) sum+=Dfs(pos+1,0,1,limit&&(i==up),s);
        else if (i && lead) sum+=Dfs(pos+1,i,0,limit&&(i==up),s+1);
        else sum+=Dfs(pos+1,i,0,limit&&(i==up),s+(i==1));
    }
    return (!lead && !limit)?f[pos][s]=sum:sum;
}
 
IL LL Cal(LL x){
    LL Ans=1,sum=0,i=0;
    len=0;
    while (x){
        z[++len]=x%2;   x/=2;
    }
    for (i=1;i<=len;i++){
        memset(f,-1,sizeof(f));
        l=i;
        sum=Dfs(1,0,1,1,0);
        Ans=(Ans*Mi(l,sum))%MOD;
    }
    return Ans;
}
 
int main(){
    #ifdef __Marvolo
    freopen("zht.in","r",stdin);
    freopen("zht.out","w",stdout);
    #endif
    scanf("%lld",&n);
    cout<<Cal(n)<<endl;
    return 0;
}

枚举+数位DP式的写法,状态的描述和传统的题目有一点区别