HDU 3652 B-number（数位DP）

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

 

 

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

 

 

Author

wqb0039

 

 

Source

2010 Asia Regional Chengdu Site —— Online Contest

 

 

 

题意：

求0 到n中数字含有13且能被13整除的数目！

PS：

数位DP模板题！

取余时用秦九韶算法！

代码如下：

 

#include 
#include 
#include 
#include 
using namespace std;
#define LL __int64
LL n, dp[25][13][3];
//dp[i][mod][j]:长度为i，余数为mod,状态为j,
int digit[25];
//nstatus: 0：不含13, 1：不含13但末尾是1, 2 :含13
LL DFS(int pos, int mod, int status, int limit)
{
    if(pos <= 0) // 如果到了已经枚举了最后一位，并且在枚举的过程中有49序列出现
        return status==2 && mod==0;//注意是 ==
    if(!limit && dp[pos][mod][status]!=-1)   //对于有限制的询问我们是不能够记忆化的
        return dp[pos][mod][status];
    LL ans = 0;
    int End = limit?digit[pos]:9;   // 确定这一位的上限是多少
    for(int i = 0; i <= End; i++)   // 每一位有这么多的选择
    {
        int nmod = (mod*10+i)%13;
        int nstatus = status;       // 有点else s = statu 的意思
        if(status==0 && i==1)//高位不含13，并且末尾不是1，现在末尾添1返回1状态
            nstatus = 1;
        else if(status==1 && i!=1 && i!=3)//高位不含13，且末尾是1，现在末尾添加的不是1返回0状态
            nstatus = 0;
        else if(status==1 && i==3)//高位不含13，且末尾是1，现在末尾添加3返回2状态
            nstatus = 2;
        ans+=DFS(pos-1, nmod, nstatus, limit && i==End);
    }
    if(!limit)
        dp[pos][mod][status]=ans;
    return ans;
}

int cal(LL x)
{
    int cnt = 0;
    while(x)
    {
        digit[++cnt] = x%10;
        x/=10;
    }
    digit[cnt+1] = 0;
    return cnt;
}

int main()
{
    while(~scanf("%I64d",&n))
    {
        memset(dp,-1,sizeof(dp));
        int len = cal(n);
        LL ans = DFS(len, 0, 0, 1);
        printf("%I64d\n",ans);
    }
    return 0;
}