Balanced Numbers數位dp

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三進位搞下， 0  表示沒出現過，  第i位為1 表示 i出現了奇數次，  2表示i 出現了偶數次。

#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;}