V

Jimmy writes down the decimal representations of all natural numbers between and including m and n, (m ≤ n). How many zeroes will he write down?

Input

Input starts with an integer T (≤ 11000), denoting the number of test cases.

Each case contains two unsigned 32-bit integers m and n, (m ≤ n).

Output

For each case, print the case number and the number of zeroes written down by Jimmy.

Sample Input

5

10 11

100 200

0 500

1234567890 2345678901

0 4294967295

Sample Output

Case 1: 1

Case 2: 22

Case 3: 92

Case 4: 987654304

Case 5: 3825876150

      让求的是【l，r】内数中含有的0的个数。直接数位dp，加上判断就可以了。

代码如下：

#include<iostream>#include<cstring>using namespace std;typedef long long LL;int bit[20];LL dp[20][20];//         数位    0的个数    前导0      上界LL dfs(int pos, int sta, bool lead, bool limit) {if (pos == -1) {if (lead)return 1;return sta;} if (!lead && !limit && dp[pos][sta] != -1)return dp[pos][sta];int up = limit ? bit[pos] : 9;LL temp = 0;for (int i = 0; i <= up; i++){if (lead == 1)temp += dfs(pos - 1, 0, i == 0, limit && i == up);//当前还是不是0else{if (i == 0)temp += dfs(pos - 1, sta + 1, 0, limit && i == up);//非前导零的0要把0数加1elsetemp += dfs(pos - 1, sta, 0, limit && i == up);}}if (!limit && !lead)dp[pos][sta] = temp;return temp;}LL solve(LL x){int pos = 0;while (x){bit[pos++] = x % 10;x /= 10;}return dfs(pos - 1, 0, 1, 1);}int main(){int t;LL le, ri;cin >> t;int cnt = 1;while (t--){memset(dp, -1, sizeof(dp));cin >> le >> ri;cout << "Case " << cnt++ << ": ";cout << solve(ri) - solve(le - 1) << endl;}return 0;}