$$\begin{gathered} \sum _{i=1}^n(pre+suc{)}^2 \\ \sum _{i=1}^{n}pr{e}^2+su{c}^2+2\times pre\times suc \\ n\times pr{e}^2+\sum su{c}^2+2\times pre\sum suc \end{gathered}$$
所以考虑维护$cnt,sum,su{m}^2$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;
}struct Res {int cnt, sum1, sum2;
}f[20][10][10];int num[20], p[20], cnt;Res dfs(int pos, int sum, int remain, int flag) {if (!pos) {if (sum != 0 && remain != 0) {return {1, 0, 0};}else {return {0, 0, 0};}}if (!flag && f[pos][sum][remain].cnt != -1) {return f[pos][sum][remain];}Res ans = {0, 0, 0};int nex = flag ? num[pos] : 9;for (int i = 0; i <= nex; i++) {if (i == 7) {continue;}Res cur = dfs(pos - 1, (sum + i) % 7, (remain * 10 + i) % 7, i == nex && flag);ans.cnt = add(ans.cnt, cur.cnt);int s = 1ll * i * p[pos] % mod;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));}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).sum2;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);for (int i = 0; i < 20; i++) {for (int j = 0; j < 10; j++) {for (int k = 0; k < 10; k++) {f[i][j][k].cnt = -1;}}}p[1] = 1;for (int i = 2; i < 20; i++) {p[i] = 1ll * p[i - 1] * 10 % mod;}int T;scanf("%d", &T);while (T--) {long long l, r;scanf("%lld %lld", &l, &r);printf("%lld\n", sub(calc(r), calc(l - 1)));}return 0;
}