[HDU 3709] Balanced Number

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 ll dp[20][2000][20];
 5 int dig[20];
 6 ll dfs(int dep,int sum,int zzz,int flag){
 7     if(!dep)return !sum;
 8     if(sum<0)return 0;
 9     if(!flag&&dp[dep][sum][zzz]!=-1)return dp[dep][sum][zzz];
10     int lim=flag?dig[dep]:9;
11     ll ans=0;
12     for(int i=0;i<=lim;i++)
13         ans+=dfs(dep-1,sum+(dep-zzz)*i,zzz,flag&(i==lim));
14     if(!flag)dp[dep][sum][zzz]=ans;
15     return ans;
16 }
17 ll solve(ll x){
18     int dd=0;
19     while(x)dig[++dd]=x%10,x/=10;
20     ll ans=0;
21     for(int i=dd;i;i--)ans+=dfs(dd,0,i,1);
22     return ans-dd+1;//0满足所有，但只算一次
23 }
24 int main(){
25     memset(dp,-1,sizeof(dp));
26     int T;
27     scanf("%d",&T);
28     while(T--){
29         ll A,B;
30         scanf("%lld%lld",&A,&B);
31         printf("%lld\n",solve(B)-solve(A-1));
32     }
33     return 0;
34 }