hdu3652（数位DP）

地址：http://acm.hdu.edu.cn/showproblem.php?pid=3652

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1744    Accepted Submission(s): 966

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

 

题意：求小于等于n的含有13且是13的倍数的数字的个数。

思路：数位dp，开四维数组，第一维表示位数，第二维表示开头数字，第三维表示对13余数，第四维表示是否含有13（1表示有0表示无）。然后就是纯数位dp模板。

代码：

#include
#include
//#include
#include
#include
//#include
using namespace std;
//#define M 1000000007
int dp[20][20][20][2]= {0};
void getdp()  //打表
{
    int i,j,k,l,s=1;
    dp[0][0][0][0]=1;
    for(i=1; i<=10; i++)
    {
        for(j=0; j<=9; j++)
        {
            for(k=0; k<=9; k++)
            {
                for(l=0; l<13; l++)
                {
                    if(k==3&&j==1)
                        dp[i][j][(l+(j*s)%13)%13][1]+=dp[i-1][k][l][0];
                    else dp[i][j][(l+(j*s)%13)%13][0]+=dp[i-1][k][l][0];
                    dp[i][j][(l+(j*s)%13)%13][1]+=dp[i-1][k][l][1];
                }
            }
        }
        s*=10;
    }
}
int main()
{
    int m,i,j,l;
    getdp();
    while(scanf("%d",&m)>0)
    {
        int len=1,len1[20]= {0},x=1,s=m,cg=0;
        if(m%13==0) cg=1;
        while(m)
        {
            len1[len++]=m%10;
            m/=10;
            x*=10;
        }
        int ans=0,bg=0;
        for(i=len-1; i>=1; i--)
        {
            for(j=0; j