Balanced Numbers数位dp

三进制搞下， 0  表示没出现过，  第i位为1 表示 i出现了奇数次，  2表示i 出现了偶数次。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include<math.h>
using namespace std;
typedef long long LL;
LL dp[20][101111];
int up[11111];
int judge(int x)
{
    int ans=0; int a[10];
    memset(a,0,sizeof(a));
    while(x){
        a[ans++]=x%3;
        x/=3;
    }
    for(int i = 0 ;i< ans;i++){
        if(a[i]==1&&(i&1)) return 0 ;
        if(a[i]==2&&!(i&1)) return 0;
    }
    return 1;
}
int change(int x,int i)
{
    int ans=0 ;int a[10];
    memset(a,0,sizeof(a));
    while(x){
        a[ans++]=x%3; x/=3;
    }
    if(a[i]==0) a[i]=1;
    else
    if(a[i]==1) a[i]=2;
    else
    if(a[i]==2) a[i]=1;
    int ans1=0;
    for(int i = 9;i>=0;i--)
        ans1=ans1*3+a[i];
    return ans1;
}

LL gao(int now,int gaojici,int first,int flag)
{
    if(now<=0) return judge(gaojici);
    if(!flag&&~dp[now][gaojici]) return dp[now][gaojici];
    LL limit = flag? up[now]: 9,ret=0;
    for(LL i= 0;i<=limit;i++){
        LL kk=change(gaojici,i);
        ret+=gao(now-1,(first||i)?kk:0,first||i,flag&&limit==i);
    }
    return flag? ret: dp[now][gaojici]=ret;
}
LL solve(LL x)
{
    int  len=0;
    while(x){
        up[++len]= x%10;
        x/=10;
    }
    return gao(len,0,0,1);
}
int main()
{
    int Icase;LL  b;LL a;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&Icase);
    while(Icase--){
        cin>>a>>b;
        cout<<solve(b)-solve(a-1)<<endl;
    }
    return 0;
}