Balanced Number 数位dp

题意：　　给出求ab之间有多少个平衡数   4139为平衡数   以3为轴   1*1+4*2==9*1

思路很好想但是一直wa  ：

注意要减去前导零的情况 0 00 000 0000   不能反复计算

#include<bits/stdc++.h>

using namespace std;

//input by bxd

#define rep(i,a,b) for(int i=(a);i<=(b);i++)

#define repp(i,a,b) for(int i=(a);i>=(b);--i)

#define RI(n) scanf("%d",&(n))

#define RII(n,m) scanf("%d%d",&n,&m)

#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)

#define RS(s) scanf("%s",s);

#define ll long long

#define REP(i,N)  for(int i=0;i<(N);i++)

#define CLR(A,v)  memset(A,v,sizeof A)

//////////////////////////////////

#define inf 0x3f3f3f3f

#define N 3700+5

#define MID N/2

ll dp[][][N];

ll a[];

ll dfs(int pos,int t,int state,bool lead,bool limit  )

{

    if(!pos)return state==;

    if(!limit&&!lead&&dp[pos][t][state+MID]!=-)return dp[pos][t][state+MID];

    ll ans=;

    int up=limit?a[pos]:;

    rep(i,,up)

    {

        ans+=dfs(pos-,t,state+i*(pos-t), lead&&i==,limit&&i==a[pos]);

    }

    if(!limit&&!lead)dp[pos][t][state+MID]=ans;

    return ans;

}

ll solve(ll x)

{

    int pos=;

    ll ans=;

    while(x)

    {

        a[++pos]=x%;

        x/=;

    }

    rep(i,,pos)

    ans+=dfs(pos,i,,true,true);

    return ans-pos+;//去除前导零

}

int main()

{

    int cas;

    CLR(dp,-);

    RI(cas);

    while(cas--)

    {

        ll a,b;

        cin>>a>>b;

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

    }

    return ;

}

或者

#include<bits/stdc++.h>

using namespace std;

//input by bxd

#define rep(i,a,b) for(int i=(a);i<=(b);i++)

#define repp(i,a,b) for(int i=(a);i>=(b);--i)

#define RI(n) scanf("%d",&(n))

#define RII(n,m) scanf("%d%d",&n,&m)

#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)

#define RS(s) scanf("%s",s);

#define ll long long

#define REP(i,N)  for(int i=0;i<(N);i++)

#define CLR(A,v)  memset(A,v,sizeof A)

//////////////////////////////////

#define inf 0x3f3f3f3f

#define N 3700+5

#define MID N/2

ll dp[][][N];

ll a[];

ll dfs(int pos,int t,int state,bool lead,bool limit  )

{

    if(!pos)return !lead&&state==;

    if(!limit&&!lead&&dp[pos][t][state+MID]!=-)return dp[pos][t][state+MID];

    ll ans=;

    int up=limit?a[pos]:;

    rep(i,,up)

    {

        ans+=dfs(pos-,t,state+i*(pos-t), lead&&i==,limit&&i==a[pos]);

    }

    if(!limit&&!lead)dp[pos][t][state+MID]=ans;

    return ans;

}

ll solve(ll x)

{

    if(x<)return ;

    if(x==)return ;

    int pos=;

    ll ans=;

    while(x)

    {

        a[++pos]=x%;

        x/=;

    }

    rep(i,,pos)

    ans+=dfs(pos,i,,true,true);

    return ans+;//去除前导零

}

int main()

{

    int cas;

    CLR(dp,-);

    RI(cas);

    while(cas--)

    {

        ll a,b;

        cin>>a>>b;

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

    }

    return ;

}