hdu 4722 Good Numbers 规律 数位dp

#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/10*10;i<=n;i++){
        ll sum = 0,tmp = i;
        while(tmp){
            sum += tmp%10;
            tmp /= 10;
        }
        if(sum%10 == 0){
            return 1;
        }
    }
    return 0;
}
ll f(ll n){
    if(ok(n)){
        return (n/10) + 1;
    }
    return n/10;
}
int main(){
    int T;
    cin >> T;
    for(int t=1;t<=T;t++){
        ll n,m;
        cin >> n >> m;
        printf("Case #%d: %lld\n",t,f(m)-f(n-1));
    }
    return 0;
}

 

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

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

 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）