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 ;
}