HDU-3652 B-number

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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
  
  

 

————————————————————集训18.2的分割线————————————————————

前言：有电脑上不上网，重启路由器却把Roll神的网顶掉了。。把Roll神气得回家了。。Sorry and Sad

思路：和不要62那道题有一定的相似度。这次的状态是：

1. 长度

2. 是否含有"13"

3. 前一位是否为1

4. 是否除得尽13

5. 是否是边界

如果已经有了"13"，向下传递的都是true。关于能否除尽13要用到同余定理的推论吧、余数和余数乘以10加k同余。

dp[len][pre][B][mod]分别是上述状态。关于是否是边界、因为是边界就不会进行记忆化，所以不需要增加维度。

代码如下：

/*
ID: j.sure.1
PROG:
LANG: C++
*/
/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <climits>
#include <iostream>
#define INF 0x3f3f3f3f
using namespace std;
/****************************************/
int num[10];
int dp[10][2][2][20];

int dfs(int len, bool pre, bool B, int mod, bool bnd)
{
	if(len == 0) {
		if(B && mod % 13 == 0) {
			//printf("who is %d\n", mod);
			return 1;
		}
		return 0;
	}	 
	if(!bnd && dp[len][pre][B][mod] != -1)
		return dp[len][pre][B][mod];
	int lim = (bnd ? num[len] : 9);
	int ret = 0;
	for(int k = lim; k >= 0; k--) {
		int nm = (mod*10+k) % 13;
		ret += dfs(len-1, k==1, B||(pre&&k==3), nm, bnd&&k==lim);
	}
	if(!bnd) dp[len][pre][B][mod] = ret;
	return ret;
}

int Solve(int x)
{
	int cnt = 0;
	while(x) {
		num[++cnt] = x % 10;
		x /= 10;
	}
	return dfs(cnt, false, false, 0, true);
}

int main()
{
#ifdef J_Sure
//	freopen("000.in", "r", stdin);
//	freopen(".out", "w", stdout);
#endif
	int n;
	memset(dp, -1, sizeof(dp));
	while(~scanf("%d", &n)) {
		printf("%d\n", Solve(n));
	}
	return 0;
}