Lightoj1068——Investigation（数位dp）

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

Output for Sample Input

Case 1: 20

Case 2: 5

Case 3: 64

两个参数，一个表示每位数之和，另一表示该数除以k的结果

WA了几次，发现数组开小了

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <cmath>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 10000005
#define Mod 10001
using namespace std;
int dight[40];
long long dp[40][100][100],k;
long long dfs(int pos,int s,bool limit,int mod)
{
if(pos==0)
return mod==0&&(s%k==0)&&s>=1;
if(!limit&&dp[pos][s][mod]!=-1)
return dp[pos][s][mod];
int end;
long long ret=0;
if(limit)
end=dight[pos];
else
end=9;
for(int d=0; d<=end; ++d)
{
int ss=s+d;
int nmod=(mod*10+d)%k;
ret+=dfs(pos-1,ss,limit&&d==end,nmod);
}
if(!limit)
dp[pos][s][mod]=ret;
return ret;
}
long long solve(long long a)
{
memset(dight,0,sizeof(dight));
int cnt=1;
while(a!=0)
{
dight[cnt++]=a%10;
a/=10;
}
return dfs(cnt-1,0,1,0);
}
int main()
{
int t,cnt=1;
scanf("%d",&t);
while(t--)
{
memset(dp,-1,sizeof(dp));
long long x,y;
scanf("%lld%lld%lld",&x,&y,&k);
printf("Case %d: %lld\n",cnt++,solve(y)-solve(x-1));
}
return 0;
}