#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define maxn 100050
int ok(ll n){
for(ll i=n/*;i<=n;i++){
ll sum = ,tmp = i;
while(tmp){
sum += tmp%;
tmp /= ;
}
if(sum% == ){
return ;
}
}
return ;
}
ll f(ll n){
if(ok(n)){
return (n/) + ;
}
return n/;
}
int main(){
int T;
cin >> T;
for(int t=;t<=T;t++){
ll n,m;
cin >> n >> m;
printf("Case #%d: %lld\n",t,f(m)-f(n-));
}
return ;
}
Good Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5193 Accepted Submission(s): 1642
Problem Description
If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number. You are required to count the number of good numbers in the range from A to B, inclusive.
Input
The first line has a number T (T <= 10000) , indicating the number of test cases. Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 1018).
Output
For test case X, output "Case #X: " first, then output the number of good numbers in a single line.
Sample Input
2
1 10
1 20
Sample Output
Case #1: 0
Case #2: 1
Hint
The answer maybe very large, we recommend you to use long long instead of int.
Source
发现： 0-10 1
0-100 10
0-1000 100
0-990 99
0-992 100
0-997 100
基本规律为 n/10 + (1或0)
加1的情况为：n/10*10 到 n 有满足条件的 比如：997: 99 + （990到997是否有满足条件的，如果有则加1）