Investigation LightOJ - 1068

Posted 2020-10-06 啦啦啦

An integer is divisible by 3 if the sum of its digits is also divisible by 3. For example, 3702 is divisible by 3 and 12 (3+7+0+2) is also divisible by 3. This property also holds for the integer 9.

In this problem, we will investigate this property for other integers.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains three positive integers A, B and K (1 ≤ A ≤ B < 2³¹ and 0 < K < 10000).

Output

For each case, output the case number and the number of integers in the range [A, B] which are divisible by K and the sum of its digits is also divisible by K.

Sample Input

3

1 20 1

1 20 2

1 1000 4

Sample Output

Case 1: 20

Case 2: 5

Case 3: 64

题解：dp[ pos ][ res1 ][ res2 ]表示当前位置，数位之和模K的余数，这些数位组成的数模K的余数。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 const int INF=0x3f3f3f3f;
 8 
 9 int L,R,K;
10 int t[10],num[10],dp[10][90][90];
11 
12 void Inite(){
13     t[0]=1;
14     for(int i=1;i<10;i++) t[i]=t[i-1]*10;
15 }
16 
17 int DFS(int pos,int res1,int res2,bool F){
18     if(pos==-1){
19         if(res1==0&&res2==0) return 1;
20         else return 0;
21     }
22     
23     if(!F&&dp[pos][res1][res2]!=INF) return dp[pos][res1][res2];
24     int maxv=F?num[pos]:9;
25     
26     int ans=0;
27     for(int i=0;i<=maxv;i++){
28         int x=(res1+i*t[pos])%K;
29         int y=(res2+i)%K;
30         ans=ans+DFS(pos-1,x,y,(F&&i==maxv)?true:false);
31     }
32     
33     if(!F) dp[pos][res1][res2]=ans;
34     return ans;
35 }
36 
37 int Solve(int temp){
38     if(temp==0) return 1;
39     int cnt=0;
40     while(temp){
41         num[cnt++]=temp%10;
42         temp/=10;
43     }
44     return DFS(cnt-1,0,0,true);
45 }
46 
47 int main()
48 {   Inite();
49 
50     int kase;
51     cin>>kase;
52     for(int k=1;k<=kase;k++){
53         cin>>L>>R>>K;
54         memset(dp,0x3f,sizeof(dp));
55         
56         int ans;
57         if(K>82) ans=0;
58         else ans=Solve(R)-Solve(L-1);
59         printf("Case %d: %d\n",k,ans);    
60     }
61     return 0;
62 }