这是一篇用递推方式写的数位DP，大家可以借鉴参考~

#include <bits/stdc++.h>
#define LL long long
const int N = 20;
const LL mod = 1000000007;
int T;
LL L, R, nums[N], cnt, f1[N][7][7][2][2], f2[N][7][7][2][2], f3[N][7][7][2][2]; 
LL Solve(LL n)
{
    cnt = 0; memset(nums, 0, sizeof(nums));
    memset(f1, 0, sizeof(f1)); memset(f2, 0, sizeof(f2)); memset(f3, 0, sizeof(f3));
    while(n)
    {
        nums[++cnt] = n % 10;
        n /= 10;
    }
    std:: reverse(nums + 1, nums + cnt + 1);
    f1[0][0][0][0][1] = 1; 
    for(int i=0;i<cnt;i++)
    {
        for(int j=0;j<7;j++)
        {
            for(int k=0;k<7;k++)
            {
                for(int l=0;l<=1;l++)
                {
                    for(int r=0;r<=1;r++)
                    {
                        if(f1[i][j][k][l][r] || f2[i][j][k][l][r] || f3[i][j][k][l][r])
                        {
                            int up = (r == 0) ? 9 : nums[i + 1];
                            for(int p=0;p<=up;p++)
                            {
                                f1[i + 1][(10 * j + p) % 7][(k + p) % 7][l || (p == 7)][r && (p == up)]
                                = (f1[i + 1][(10 * j + p) % 7][(k + p) % 7][l || (p == 7)][r && (p == up)] + f1[i][j][k][l][r]) % mod;
                                f2[i + 1][(10 * j + p) % 7][(k + p) % 7][l || (p == 7)][r && (p == up)]
                                = (f2[i + 1][(10 * j + p) % 7][(k + p) % 7][l || (p == 7)][r && (p == up)] + 10 * f2[i][j][k][l][r] + p * f1[i][j][k][l][r]) % mod;
                                f3[i + 1][(10 * j + p) % 7][(k + p) % 7][l || (p == 7)][r && (p == up)]
                                = (f3[i + 1][(10 * j + p) % 7][(k + p) % 7][l || (p == 7)][r && (p == up)] + 100 * f3[i][j][k][l][r] + 20 * p * f2[i][j][k][l][r] + p * p * f1[i][j][k][l][r]) % mod;
                            }
                        }
                    }
                }
            }
        }
    }
    LL res = 0;
    for(int j=1;j<7;j++)
        for(int k=1;k<7;k++)
            for(int r=0;r<=1;r++)
            {
                res = (res + f3[cnt][j][k][0][r]) % mod;
            }
    return res;
}
int main()
{
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lld%lld", &L, &R);
        printf("%lld\n", (Solve(R) - Solve(L - 1) + mod) % mod);
    }
    return 0;
}