LightOJ1068(数位DP)

最新推荐文章于 2024-08-29 12:01:12 发布

qkoqhh

最新推荐文章于 2024-08-29 12:01:12 发布

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分类专栏： DP

本文链接：https://blog.youkuaiyun.com/qkoqhh/article/details/79249966

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本文探讨了使用数位动态规划解决一类特殊整数问题的方法。这类问题要求找出指定范围内，既可被特定整数K整除且其各位数字之和也能被K整除的整数数量。文章通过实例讲解了如何有效利用数位DP进行状态压缩和转移，避免内存溢出等问题。

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发现数位DP其实并不难调。。只要有数据的话。。。（当然考场需要对拍。。

这个要维护2个余数，然后就是按照数位该有的模式转移即可，不过k的范围非常大，貌似会爆内存。。以前做数位dp其实根本就没想到会到内存不够用的地步丫。。仔细想想会发现，数位和最多也才82（1999999999的情况），所以k>82直接输出0就可以了，然后这样内存还是非常充裕的。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define eps 1e-8
#define inf 1e9
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define ls T[i<<1]
#define rs T[i<<1|1]
#define op T[i]
#define mid (x+y>>1)
#define NM 83
#define nm 100498
#define pi 3.141592653
using namespace std;
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return f*x;
}






int _x,_y,k,n,b[NM],d[11][NM][NM];
bool V[11][NM][NM];
int dfs(int i,bool f,int t,int v){
	if(!i)return !t&&!v;
	if(!f&&V[i][t][v])return d[i][t][v];
	int m=f?b[i]:9,ans=0;
	inc(j,0,m)ans+=dfs(i-1,f&&j==m,(t*10+j)%k,(v+j)%k);
	if(!f)d[i][t][v]=ans,V[i][t][v]++;
	return ans;
}

int solve(int x){
	n=0;
	for(int t=x;t;t/=10)b[++n]=t%10;
	return dfs(n,1,0,0);
}

int main(){
	int _=read();
	inc(T,1,_){
		_x=read();_y=read();k=read();
		if(k>82){printf("Case %d: 0\n",T);continue;}
		mem(V);mem(d);
		printf("Case %d: %d\n",T,solve(_y)-solve(_x-1));
	}
	return 0;
}

Investigation

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

[Submit] [Go Back] [Status]

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

1 20 1

1 20 2

1 1000 4

Sample Output

Case 1: 20

Case 2: 5

Case 3: 64

Source

Problem Setter: Sohel Hafiz

Special Thanks: Jane Alam Jan (Solution, Dataset)