hdu 3652 B-number

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2266    Accepted Submission(s): 1227

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

求从1~n包含13且能被13整除的数的个数，判断包含13保存上一位，判断整除13保存余数。

#include 
#include 
#include 
using namespace std;

#define LL long long
int p[20];
LL dp[20][15][15][3];
LL dfs(int h,int last,int m,bool ok,bool sign){
	if(h<0) {
		if(m==0 && ok) return 1;
		return 0;
	}
	if(!sign && dp[h][last][m][ok]!=-1) return dp[h][last][m][ok];
	int end = sign?p[h]:9;
	LL ans = 0;
	for(int i=0;i<=end;i++){
		if(last==1 && i==3) ans += dfs(h-1,i,(m*10+i)%13,true,sign && i==p[h]);
		else ans += dfs(h-1,i,(m*10+i)%13,ok,sign && i==p[h]);
	}
	if(!sign) dp[h][last][m][ok] = ans;
	return ans;
}
LL solve(LL n){
	int cnt = 0;
	while(n>0){
		p[cnt++] = n%10;
		n /= 10;
	}
	return dfs(cnt-1,0,0,false,true);
}
int main(){
	LL n;
	memset(dp,-1,sizeof dp);
	while(cin >>n){
		cout << solve(n) <