Lightoj 1068 数位dp

Investigation

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Description

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 byK.

Sample Input

3

1 20 1

1 20 2

1 1000 4

Sample Output

Case 1: 20

Case 2: 5

Case 3: 64

题意：a和b，求出a和b区间能被k整除且它各个位的和也能被k整除的数的个数。

题解：数位dp。

#include <stdio.h>  
#include <string.h>  
#include <algorithm>  
using namespace std;  
typedef long long ll;
ll bit[15];  
ll dp[10][123][105][2],k;  
ll dfs(ll pos,ll mod,ll have,ll lim)
{  
    ll num,i,ans,mod_x,have_x;  
    if(pos<=0)  
        return mod%k==0&&have%k==0;  
    if(!lim&&dp[pos][mod][have][lim]!=-1)
        return dp[pos][mod][have][lim];  
    num=lim?bit[pos]:9;
    ans=0;  
    for(i=0;i<=num;i++){  
        ans+=dfs(pos-1,mod+i,(have*10+i)%k,lim&&i==num);
    }  
    if(!lim)  
        dp[pos][mod][have][lim]=ans;  
    return ans;  
}  
  
int main(){  
    ll n,len,t,m,cas=1,i;
	scanf("%lld",&t); 
    while(t--){  
        len=0;
        scanf("%lld%lld%lld",&n,&m,&k);
		if(k>=100){
			printf("Case %lld: %lld\n",cas++,0);
			continue;
		}
	 	memset(dp,-1,sizeof(dp));
        memset(bit,0,sizeof(bit));   
		n--;
        while(n){  
            bit[++len]=n%10;  
            n/=10;  
        }  
        ll sol1=dfs(len,0,0,1);
        len=0;
		while(m){
			bit[++len]=m%10;
			m/=10;
		}
		ll sol2=dfs(len,0,0,1);
		printf("Case %lld: %lld\n",cas++,sol2-sol1);
    }  
    return 0;  
}