Codeforces Round #582 (Div. 3) C. Book Reading

Portal

Meaning of the questions:

You n, k. It is shown in [1, n] within this range, to find out all within the interval x satisfies x% k == 0, then let x all bits are added together (i.e., x% 10), output.

For example: Input 102

The number is then meet 246810

Then 2% 4% 10 + 10 + 10 + 8% 6% 10% 10 = 10 + 20

Then output 20

 

answer:

Satisfying x% k == x 0, it is certainly a factor of k and the product of y, i.e., 1 * k, 2 * k, 3 * k, 4 * k ....... (y <= n / k)

We find the same bits 11 * k bits and the number of 1 * k, i.e., (11 * k)% 10 == (1 * k)% 10

Well certainly there are (21 * k)% 10 == (1 * k)% 10 ...............

Therefore, we only need to find out all (i * k)% 10 (1 <= i <= 9) and then added together

 

Then look at the y is the number, get what you can, look at the specific code!

 

Code:

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 const int maxn=15;
 7 typedef long long ll;
 8 ll v[maxn],sum[maxn];
 9 int main()
10 {
11     ll t;
12     cin>>t;
13     while(t--)
14     {
15         ll n,m;
16         cin>>n>>m;
17  
18         for(ll i=1;i<=9;++i)
19             v[i]=(i*m)%10,sum[i]=sum[i-1]+v[i];
20         if(n<m)
21         {
22             printf("0\n");
23             continue;
24         }
25         n/=m;
26         m%=10;
27         ll summ=0,ans;
28         ans=n/10;
29         ll q=n%10;
30         //printf("%d %lld %d\n",n,ans,q);
31         summ=ans*sum[9];
32         summ+=sum[q];
33         printf("%I64d\n",summ);
34     }
35     return 0;
36 }

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