BZOJ1833（数位dp）

这个数位dp倒是没什么限制条件，只是需要在过程中把每个数字出现次数记录一下即可。记忆化返回时数学算出。框架还是套板子。

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std; typedef long long ll;
 ll n, m, cnt1[], cnt2[], dp[][];
 int tot, a[]; ll dfs(int pos, int state0, int limit, ll *cnt) {
     if (!pos) {
         cnt[] += state0;
         return ;
     }
     if (!limit && !state0 && dp[pos][state0] != -) {
         for (int i = ; i <= ; i++) {
             cnt[i] += pow(, pos - ) * pos;
         }
         return dp[pos][state0];
     }     int up = limit ? a[pos] : ;
     ll ret = ;     for (int i = ; i <= up; ++i) {
         ll tmp = dfs(pos - , state0 & (!i), limit & (i == a[pos]), cnt);
         cnt[i] += tmp;
         if (i ==  && state0)    cnt[i] -= tmp;
         ret += tmp;
     }     if (!limit)    dp[pos][state0] = ret;
     return ret;
 } void solve(ll x, ll *cnt) {
     for (tot = ; x; x /= )
         a[++tot] = x % ;
     dfs(tot, , , cnt);
 } int main() {
     memset(dp, -, sizeof dp);     cin >> n >> m;
     solve(m, cnt1);
     solve(n - , cnt2);     for (int i = ; i <= ; ++i)
         printf("%lld ", cnt1[i] - cnt2[i]);     return ;
 }