[HDU_3652]B-number

题目描述

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.

输入

Process till EOF. In each line, there is one positive integer n(1 <= n <= 1000000000).

输出

Print each answer in a single line.

样例输入

13
100
200
1000

样例输出

1
1
2
2

求小于n是13的倍数且含有“13”的数的个数，有多组数据，输入到结束。

dp[i][j][k][l]。i表示i位数j表示第i位是j的k表示是否包含13（k==0 or 1）L表示膜13是l的数的个数

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int dp[][][][];
int e[];
void init()
{
    e[]=;
    for(int i=;i<=;i++)e[i]=(e[i-]*)%;
    dp[][][][]=;
    for(int i=;i<=;i++)
    for(int j1=;j1<=;j1++)
    for(int j2=;j2<=;j2++)
    for(int l=;l<;l++)
    {
        int yu=(l-(j1*e[i-])%+)%;//yu表示dp[][][][l]从dp[][][][yu]转移//yu+j1贡献的余数=l
        dp[i][j1][][l]+=dp[i-][j2][][yu];
        if(j1==&&j2==)dp[i][j1][][l]+=dp[i-][j2][][yu];
        else dp[i][j1][][l]+=dp[i-][j2][][yu];
    }
 }
int solve(int a)
{
    int len=,ans=,mod=,d[];bool t=;
    while(a)
    {
        d[++len]=a%;a/=;
    }
    d[len+]=;
    for(;len>=;len--)
    {
        for(int i=;i<d[len];i++)
        {
            int yu=(-(mod*e[len])%)%;//算上高位mod13的余数还要加多少（yu）mod13才会得0
            ans+=dp[len][i][][yu];
            if(t||d[len+]==&&i==)ans+=dp[len][i][][yu];
        }
        if(d[len+]==&&d[len]==)t=;
        mod=(mod*+d[len])%;
    }
    return ans;
}
int main()
{
    int n;init();
    while(scanf("%d",&n)!=EOF)
    {
        printf("%d\n",solve(n+));
    }
    return ;
}