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[10], cnt2[10], dp[20][2];
int tot, a[20];

ll dfs(int pos, int state0, int limit, ll *cnt) {
    if (!pos) {
        cnt[0] += state0;
        return 1;
    }
    if (!limit && !state0 && dp[pos][state0] != -1) {
        for (int i = 0; i <= 9; i++) {
            cnt[i] += pow(10, pos - 1) * pos;
        }
        return dp[pos][state0];
    }

    int up = limit ? a[pos] : 9;
    ll ret = 0;

    for (int i = 0; i <= up; ++i) {
        ll tmp = dfs(pos - 1, state0 & (!i), limit & (i == a[pos]), cnt);
        cnt[i] += tmp;
        if (i == 0 && state0)    cnt[i] -= tmp;
        ret += tmp;
    }

    if (!limit)    dp[pos][state0] = ret;
    return ret;
}

void solve(ll x, ll *cnt) {
    for (tot = 0; x; x /= 10)
        a[++tot] = x % 10;
    dfs(tot, 1, 1, cnt);
}

int main() {
    memset(dp, -1, sizeof dp);

    cin >> n >> m;
    solve(m, cnt1);
    solve(n - 1, cnt2);

    for (int i = 0; i <= 9; ++i)
        printf("%lld ", cnt1[i] - cnt2[i]);
    
    return 0;
}