HDU - 3652 B-number【数位DP】

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Time limit 1000 ms
Memory limit 32768 kB

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.

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题目分析

B-number B数，比较套路的入门数位DP

$dp[i][mod][st]$表示 当前还剩 $i$位没填，除以13余数为mod
st=0表示13还没出现过；st=1表示13没出现过且第 $i+1$位为1；st=2表示13出现过
的满足条件的数字个数

套上数位DP的记搜板就好了

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#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
 
int read()
{
    int f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}

const int maxn=50;
int n;
int dp[maxn][15][5],dig[maxn];

int qpow(int a,int k,int p)
{
	int res=1;
	while(k>0){
		if(k&1) res=(res*a)%p;
		a=(a*a)%p; k>>=1;
	}
	return res;
}

int DP(int len,int mod,int st,int pre)
{
	if(len==0) return (mod==0&&st==2);
	if(!pre&&dp[len][mod][st]!=-1) return dp[len][mod][st];
	
	int res=0,mx=pre?dig[len]:9;
	for(int i=0;i<=mx;++i)
	{	
		int nxt;
		if(st==2||(st==1&&i==3)) nxt=2;
		else if((st==0||st==1)&&i==1) nxt=1;
		else nxt=0;
		res+=DP(len-1,(mod+qpow(10,len,13)*i)%13,nxt,pre&&i==mx);
	}
	if(!pre) dp[len][mod][st]=res;
	return res;
}

int solve(int x)
{
	int len=0;
	while(x)
	{
		dig[++len]=x%10;
		x/=10;
	}
	return DP(len,0,0,1);
}

int main()
{
    while(scanf("%d",&n)!=EOF)
	{
		memset(dp,-1,sizeof(dp));
		printf("%d\n",solve(n));
	}
	return 0;
}