不要666升级版（数位DP，三次方和）

不要666升级版

$∑(pre+suc)2n×pre2+2×pre∑suc+∑suc2\sum(pre + suc) ^ 2\\ n \times pre ^ 2 + 2 \times pre \sum suc + \sum suc ^ 2\\$

$∑(pre+suc)3∑(pre3+3×pre2×suc+3×pre×suc2+suc3)n×pre3+3×pre2∑suc+3×pre∑suc2+∑suc3\sum (pre + suc) ^ 3\\ \sum \left(pre ^ 3 + 3 \times pre ^ 2 \times suc + 3 \times pre \times suc ^ 2 + suc ^ 3\right)\\ n \times pre ^ 3 + 3 \times pre ^ 2 \sum suc + 3 \times pre \sum suc ^ 2 + \sum suc ^ 3\\$

由此维护 $n, sum, sum^ 2, sum ^ 3$ 即可，定义 $f [i] [j] [k]$ ，为枚举到第 $i$ 为，数位和为 $j$ ，数字余 $k$ 的状态，记忆化搜索即可。

#include <bits/stdc++.h>using namespace std;const int mod = 1e9 + 7;inline int add(int x, int y) {return x + y < mod ? x + y : x + y - mod;
}inline int sub(int x, int y) {return x >= y ? x - y : x - y + mod;
}int cnt, num[20], p[20];struct Res {int cnt = -1, sum1, sum2, sum3;
}f[20][10][10];Res dfs(int pos, int sum, int remain, int flag) {if (!pos) {if (sum != 0 && remain != 0) {return {1, 0, 0, 0};}else {return {0, 0, 0, 0};}}if (!flag && f[pos][sum][remain].cnt != -1) {return f[pos][sum][remain];}Res ans = {0, 0, 0, 0};int nex = flag ? num[pos] : 9;for (int i = 0; i <= nex; i++) {if (i == 6) {continue;}Res cur = dfs(pos - 1, (sum + i) % 6, (remain * 10 + i) % 6, i == nex && flag);int s = 1ll * i * p[pos] % mod;ans.cnt = add(ans.cnt, cur.cnt);ans.sum1 = add(ans.sum1, add(1ll * s * cur.cnt % mod, cur.sum1));ans.sum2 = add(add(ans.sum2, 1ll * cur.cnt * s % mod * s % mod), add(cur.sum2, 2ll * s % mod * cur.sum1 % mod));ans.sum3 = add(ans.sum3, add(add(1ll * cur.cnt * s % mod * s % mod * s % mod, 3ll * s % mod * s % mod * cur.sum1 % mod), add(3ll * s % mod * cur.sum2 % mod, cur.sum3)));}if (!flag) {f[pos][sum][remain] = ans;}return ans;
}int calc(long long x) {cnt = 0;while (x) {num[++cnt] = x % 10;x /= 10;}return dfs(cnt, 0, 0, 1).sum3;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);p[1] = 1;for (int i = 2; i < 20; i++) {p[i] = 1ll * p[i - 1] * 10 % mod;}long long l, r;while (scanf("%lld %lld", &l, &r) != EOF) {printf("%lld\n", sub(calc(r), calc(l - 1)));}return 0;
}