数位dp-HDU3652

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

131002001000

Sample Output

1122

             题目：在一个数的范围里找一种能被13整除且数字中含有13的数

             dp[]i[j]k[]，i是数字位数， j是模13的余数，k是状态的

    

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int a[20];int dp[20][20][3];//状态0：前一位不为1，状态1：前一位为1，状态2：前一位为1当前位为3int dfs(int pos, int mod, int sta, bool limit){    if(pos==0)        return (mod==0&&sta==2);    if(!limit && dp[pos][mod][sta]!= -1)        return dp[pos][mod][sta];    int up = limit ? a[pos]:9;    int temp=0;    for(int i=0;i<=up;i++)    {        int newmod = (mod*10 + i)%13;        int cc = sta;        if(sta==0 && i==1)cc=1;        if(sta==1 && i!=1)cc=0;        if(sta==1 && i==3)cc=2;        temp += dfs(pos-1,newmod, cc, limit && i==up);       // printf("%d\n", temp);    }    if(!limit)        dp[pos][mod][sta] = temp;    return temp;}int solve(int x){    int len = 0;    while(x)    {        a[++len] =  x%10;        x/=10;    }    return dfs(len, 0, 0, true);}int main(){    int n;    memset(dp, -1, sizeof(dp));    while(~scanf("%d",&n))    {        printf("%d\n", solve(n));    }    return 0;}