HDU - 3652 B-number

最新推荐文章于 2020-04-18 20:46:01 发布

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最新推荐文章于 2020-04-18 20:46:01 发布

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分类专栏： DP 数位dp

本文链接：https://blog.youkuaiyun.com/oWuHen12/article/details/79749861

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本文介绍了一种使用数位DP的方法来解决寻找特定范围内既包含子字符串'13'又能被13整除的非负整数数量的问题。通过深度优先搜索（DFS）策略实现了高效的计算。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3652

题目大意：1~n 中包含13且是13的倍数的数的个数

sol：数位dp模版题

dfs(pos,pre,mo,ok,limit) 表示正在处理第pos位，前一位为pre（用于处理是否含13），对13取模为mo，是否已经包含ok，是否满足边界限制limit的数的个数。

B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7852 Accepted Submission(s): 4629

Problem Description

A wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the sub- string "13" and can be divided by 13. For example, 130 and 2613 are wqb-numbers, but 143 and 2639 are not. Your task is to calculate how many wqb-numbers from 1 to n for a given integer n.

Input

Process till EOF. In each line, there is one positive integer n(1 <= n <= 1000000000).

Output

Print each answer in a single line.

Sample Input

  13 100 200 1000
 

Sample Output

  1 1 2 2
 

code:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>

using namespace std;

typedef long long ll;
const int maxn = 60;
const int mod = 1e9+7;

ll dp[40][10][13][2];
int a[maxn];


ll dfs(int pos,int pre,int mo,bool ok,bool limit){
	if(pos==-1) {
		if(ok&&mo==0) return 1;
		else return 0;
	}
	if(!limit&&dp[pos][pre][mo][ok]!=-1) return dp[pos][pre][mo][ok];
	ll ans=0;
	int up=limit?a[pos]:9;
	for(int i=0;i<=up;i++){
		if(pre==1){
			ans=ans+dfs(pos-1,i,(mo*10+i)%13,ok||(i==3),limit&&i==up);
		}else {
			ans=ans+dfs(pos-1,i,(mo*10+i)%13,ok,limit&&i==up);
		}
	}
	if(!limit) {
		dp[pos][pre][mo][ok]=ans;
	}
	return ans;
}

ll sol(ll x){
	int pos=0;
	while(x){
		a[pos++]=x%10;
		x/=10;
	}
	return dfs(pos-1,0,0,false,true);
}

int main(){
	ll n;
	memset(dp,-1,sizeof(dp));
	while(scanf("%lld",&n)==1){
		ll ans=sol(n);
		printf("%lld\n",ans);
	}
	return 0;
}