B-number HDU

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

#include <stdio.h>#include <iostream>#include <algorithm>#include <cmath>#include <map>#include <vector>using namespace std;int bit[20];int dp[20][13][3];//    位数，  前一位的mod13的余数， 状态// 状态0表示 尾数数不是 1   状态1表示 尾数为1   状态2表示   包含13int dfs(int pos, int mod, int state, bool limit) {    if (pos == -1)        return mod == 0 && state == 2;    if (!limit && dp[pos][mod][state] != -1)        return dp[pos][mod][state];    int up = limit ? bit[pos] : 9;    int temp = 0;    for (int i = 0; i <= up; ++i) {        int mod_n = (mod*10 + i) % 13;        int have = state;           //如果state == 2，则状态不需要改变        if (state == 0 && i == 1)   //上一位非1，这一位为1，改变状态            have = 1;        if (state == 1 && i != 1)   //上一位为1，这一位非1，改变状态            have = 0;        if (state == 1 && i == 3)   //上一位为1，这一位为3，令状态state = 2；            have = 2;        temp += dfs(pos-1, mod_n, have, limit && i == bit[pos]);    }        if (!limit)        dp[pos][mod][state] = temp;    return temp;}int solve(long long x) {    int i = 0;    while (x) {        bit[i++] = x%10;        x /= 10;    }    return dfs(i-1, 0, 0, 1);}int main() {    long long n;    while (cin >>n) {        memset(dp, -1, sizeof(dp));        cout << solve(n) << endl;    }}