# 数位DP入坑

Hdu 2089 不要62

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long LL;

inline int read()
{
    char c=getchar();int num=;
    for(;!isdigit(c);c=getchar());
    for(;isdigit(c);c=getchar())
        num=num*+c-'';
    return num;
}

const int N=1e6+;

int n,m;
int a[],dgt;
int f[][;

int dfs(int dep,int pre,int sta,bool lim)
{
    if(!dep)
        return ;
    if(!lim&&f[dep][sta!=-)
        return f[dep][sta];
    int up=lim?a[dep]:;
    int ans=;
    for(int i=;i<=up;++i)
    {
        if(i==) continue;
        if(pre==&&i==) continue;
        ans+=dfs(dep-,i,i==,lim&&i==a[dep]);
    }
    if(!lim)
        f[dep][sta]=ans;
    return ans;
}

int solve(int x)
{
    for(dgt=;x;a[++dgt]=x%,x/=);
    return dfs(dgt,,,);
}

int main()
{
    while()
    {
        memset(f,-,sizeof(f));
        n=read(),m=read();
        if(!n&&!m)
            break;
        cout<<solve(m)-solve(n-)<<'\n';
    }
    return ;
}

Hdu 6148 Valley Numer

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long LL;

inline int read()
{
    char c=getchar();int num=;
    for(;!isdigit(c);c=getchar());
    for(;isdigit(c);c=getchar())
        num=num*+c-'';
    return num;
}

const int N=;
const int mod=1e9+;

int T,n;
char a[N];
int f[N][][];

int dfs(int dep,int pre,int turn,bool lim,bool invalid)
{
    if(dep==n+)
        return invalid?:;
    if(!lim&&f[dep][pre][turn]!=-)
        return f[dep][pre][turn];
    int up=lim?a[dep]-='':;
    int ans=;
    for(int i=;i<=up;++i)
    {
        if(turn==&&i<pre)
            continue;
        int p=;
        if(invalid) p=;
        else if(i>pre) p=;
        else if(i<pre) p=;
        else p=turn;
        ans+=dfs(dep+,i,p,lim&&i==a[dep],invalid&&i==);
        ans%=mod;
    }
    if(!lim)
        f[dep][pre][turn]=ans;
    return ans;
}

int main()
{
    T=read();
    while(T--)
    {
        memset(f,-,sizeof(f));
        scanf("%s",a+);
        n=strlen(a+);
        cout<<dfs(,,,,)<<'\n';
    }
    return ;
}

Bzoj 1026: [SCOI2009]windy数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

typedef long long LL;

inline LL read()
{
    char c=getchar();LL num=;
    for(;!isdigit(c);c=getchar());
    for(;isdigit(c);c=getchar())
        num=num*+c-'';
    return num;
}

LL A,B;
LL a[],dgt;
LL f[][];

LL dfs(int dep,int pre,bool lim,bool invalid)
{
    if(!dep)
        return invalid?:;
    if(!lim&&!invalid&&f[dep][pre]!=-)
        return f[dep][pre];
    int up=lim?a[dep]:;
    LL ans=;
    for(int i=;i<=up;++i)
    {
        if(!invalid&&abs(i-pre)<) continue;
        ans+=dfs(dep-,i,lim&&i==a[dep],invalid&&i==);
    }
    if(!lim&&!invalid)
        f[dep][pre]=ans;
    return ans;
}

LL solve(LL x)
{
    for(dgt=;x;a[++dgt]=x%,x/=);
    return dfs(dgt,,,);
}

int main()
{
    A=read(),B=read();
    memset(f,-,sizeof(f));
    cout<<solve(B)-solve(A-);
    return ;
}

Poj 3252  Round Numbers

题意：问[l,r]中二进制位1比0少的数有多少个。

Sol: 记录当前层0,1个数。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

typedef long long LL;

inline int read()
{
    char c=getchar();int num=;
    for(;!isdigit(c);c=getchar());
    for(;isdigit(c);c=getchar())
        num=num*+c-'';
    return num;
}

int l,r;
int a[],bit;
int f[][][];

int dfs(int dep,int c0,int c1,bool lim,bool invalid)
{
    if(!dep)
        return (!invalid&&c0>=c1)?:;
    if(!lim&&f[dep][c0][c1]!=-)
        return f[dep][c0][c1];
    int up=lim?a[dep]:;
    int ans=;
    for(int i=;i<=up;++i)
        ans+=dfs(dep-,c0+(i==&&!invalid),c1+(i==),lim&&i==up,invalid&&i==);
    if(!lim)
        f[dep][c0][c1]=ans;
    return ans;
}

int solve(int x)
{
    for(bit=;x;a[++bit]=x&,x>>=);
    return dfs(bit,,,,);
}

int main()
{
    memset(f,-,sizeof(f));
    l=read(),r=read();
    cout<<solve(r)-solve(l-);
    return ;
}

P2518 [HAOI2010]计数

一个比较假的数位DP？但也是按照数位来做的。

Sol:无。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

typedef long long LL;

inline int read()
{
    char c=getchar();int num=;
    for(;!isdigit(c);c=getchar());
    for(;isdigit(c);c=getchar())
        num=num*+c-'';
    return num;
}

char num[];
LL ans;int n,m;
int a[],cnt[];

LL c[][];
void init()
{
    c[][]=;
    for(int i=;i<=;++i)
    {
        c[i][]=;
        for(int j=;j<=i;++j)
            c[i][j]=c[i-][j-]+c[i-][j];
    }
}

LL C(int n)
{
    LL res=;
    for(int i=;i<;++i)
        if(cnt[i])
            res*=c[n][cnt[i]],n-=cnt[i];
    return res;
}

int main()
{
    init();
    scanf("%s",num+);
    for(n=;num[n];++n)
        a[n]=num[n]-'',++cnt[num[n]-''];
    m=--n;
    for(int i=;i<=n;++i)
    {
        --m;
        for(int j=;j<a[i];++j)
            if(cnt[j])
                --cnt[j],ans+=C(m),++cnt[j];
        --cnt[a[i]];
    }
    cout<<ans;
    return ;
}