G-B-number

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

      给一个数n，让求1-n之间有多少个含有相邻的1、3并且能被13整除的数。解法完全是套模板了，不过第一个题，还是看了人家的才对。 

      ！！！这里的一个问题：：：dp数组的第二位mod，也就是dfs的第二个参数mod，这里指的是求得模。到最后判断是return 1 / 0 的标准是：①mod==0；②sta==2。

代码如下：

#include <iostream>#include <cstring>using namespace std;typedef long long LL;int a[20];LL dp[20][20][3];//dp[pos][mod][sta]//pos: 数位//mod: 余数//sta: 3种操作状况，//0：末尾不是1, 1：末尾是1,  2：含有13LL dfs(int pos, int mod, int sta, bool limit)//      当前位   mod   status  是否有上限{if (pos <= 0)//最后一位{if (mod == 0 && sta == 2)//没有余数 + 有13return 1;return 0;}if (!limit && dp[pos][mod][sta] != 0)//没有上限 被访问过return dp[pos][mod][sta];int up = limit ? a[pos] : 9;LL temp = 0;for (int i = 0; i <= up; i++){int mod2 = (mod * 10 + i) % 13;//int sta2 = sta;if (sta == 1 && i != 1)//末尾是1，现在加入的不是1sta2 = 0;//标记为末尾不是1if (sta == 0 && i == 1)//末尾不是1，现在加入的是1sta2 = 1;//标记为末尾是1if (sta == 1 && i == 3)//末尾是1，现在加入的是3sta2 = 2;//标记为含有13temp += dfs(pos - 1, mod2, sta2, limit && i == up);}if (!limit)dp[pos][mod][sta] = temp;return temp;}LL solve(int x){int pos = 1;while (x){a[pos++] = x % 10;x /= 10;}return dfs(pos - 1, 0, 0, 1);//          首位  mod sta  有上限}int main(){int n;while (cin >> n){memset(a, 0, sizeof(a));memset(dp, 0, sizeof(dp));cout << solve(n) << endl;}return 0;}