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;

}