2
1 10
1 20

Case #1: 0
Case #2: 1
Hint
The answer maybe very large, we recommend you to use long long instead of int.
 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

ll s[20];
ll dp[20][10];//dp[i][j]表示长度为i的数对10取模为j

ll slove(ll x)
{
    ll t=0,sum=0;
    while (x)
    {
        s[++t]=x%10;
        x/=10;
    }
    ll ans=0,m=0;
    CL(dp);
    for (int i=t; i>0; i--)//最高位开始枚举
    {
        for (int j=0; j<10; j++)//没有界限，枚举所有
            for (int k=0; k<10; k++)
            dp[i][(j+k)%10]+=dp[i+1][j];
        for (int j=0; j<s[i]; j++)//有界限，如上一位为1，该位为2；而上一位已经是1了，所以该位只能取到2
            dp[i][(j+m)%10]++;
        m = (m+s[i])%10;//保存余数
    }
    if (!m) dp[1][0]++;
    return dp[1][0];
}

int main ()
{
    int T,ii=1;
    ll a,b;
    scanf ("%d",&T);
    while (T--)
    {
        scanf ("%lld%lld",&a,&b);
        printf ("Case #%d: %lld\n",ii++,slove(b)-slove(a-1));
    }
    return 0;
}