[SPOJ BALNUM] Balanced Numbers

dp[dep][ex][sta]表示长度为dep的，前面出现过的数的集合为ex，不满足要求的数字的集合为sta的满足要求的数的个数

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 ll dp[20][1024][1024];
 5 int dig[20],tmp;
 6 ll dfs(int dep,int ex,int sta,int zero,int flag){
 7     if(!dep)return (ex&sta)==0;
 8     if(!flag&&dp[dep][ex][sta]!=-1)return dp[dep][ex][sta];
 9     int lim=flag?dig[dep]:9;
10     ll ans=0;
11     for(int i=0;i<=lim;i++){
12         if(zero&&i==0)ans+=dfs(dep-1,0,tmp,1,flag&(i==lim));
13         else ans+=dfs(dep-1,ex|(1<<i),sta^(1<<i),0,flag&(i==lim));
14     }
15     if(!flag)dp[dep][ex][sta]=ans;
16     return ans;
17 }
18 ll solve(ll x){
19     int dd=0;
20     while(x)dig[++dd]=x%10,x/=10;
21     return dfs(dd,0,tmp,1,1);
22 }
23 int main(){
24     memset(dp,-1,sizeof(dp));
25     for(int i=0;i<10;i+=2)tmp|=(1<<i);
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 }