hdu 3652 B-number 数位DP

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.

InputProcess till EOF. In each line, there is one positive integer n(1 <= n <= 1000000000).OutputPrint each answer in a single line.Sample Input

13
100
200
1000

Sample Output

1
1
2
2

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <string>
using namespace std;
typedef long long LL;
int dp[15][15], n, m, num[10];
void init() {
    dp[0][0] = 1;
    for (int i = 1 ; i <= 7 ; i++) {
        for (int j = 0 ; j < 10 ; j++) {
            for (int k = 0 ; k < 10 ; k++) {
                if (j == 4 || (j == 6 && k == 2)) continue;
                dp[i][j] += dp[i - 1][k];
            }
        }
    }
}
int getlen(int x) {
    int ret = 0;
    while(x) ret++, x /= 10;
    return ret;
}
void getnum(int x, int len) {
    memset(num, 0, sizeof(num));
    for (int i = 1 ; i <= len ; i++) {
        num[i] = x % 10;
        x /= 10;
    }
}
int getans(int x) {
    int ans = 0, len = getlen(x);
    getnum(x,len);
    for (int i = len ; i >= 1 ; i--) {
        for (int j = 0 ; j < num[i] ; j++) {
            if (j == 4 || (j == 2 && num[i + 1] == 6) ) continue;
            ans += dp[i][j];
        }
        if (num[i] == 4 || (num[i] == 2 && num[i + 1] == 6)) break;
    }
    return ans;
}
int main() {
    init();
    while(scanf("%d%d", &n, &m) != EOF) {
        if (n == 0 && m == 0) break;
        printf("%d
", getans(m + 1) - getans(n));
    }
    return 0;
}