HDU3652 B-number [数位DP]

HDU3652 B-number

题意：1到n含有13且整除13的数字个数

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(f[d][q][one][has])表示d位余数为q上一位是否为1当前是否有13到0位时合法数字个数
除了天际线全都记忆化

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N=15;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, a[N], len, f[N][N][2][2];
int dfs(int d, int q, bool one, bool has, int sky) {
	if(d==0) return q==0 && has;
	if(!sky && f[d][q][one][has]!=-1) return f[d][q][one][has];
	int lim= sky ? a[d] : 9, now=0;
	for(int i=0; i<=lim; i++) 
		now += dfs(d-1, (q*10 + i)%13, i==1, has || (one && i==3), sky && i==lim);
	return sky ? now : f[d][q][one][has]=now;
}
int cal(int n) {
	memset(f,-1,sizeof(f));
	len=0;
	while(n) a[++len]=n%10, n/=10;
	return dfs(len, 0, 0, 0, 1);
}
int main() {
	freopen("in","r",stdin);
	while(scanf("%d",&n)!=EOF) printf("%d
", cal(n));
}