与13有关（数位dp）

找出1~n范围内含有13并且能被13整除的数字的个数.#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int n, shu[20], dp[20][20][10];
int dfs(int len, int mod, int state, bool shangxian)
{
if (len == 0)
return mod == 0 && state == 2;
if (!shangxian && dp[len][mod][state])
return dp[len][mod][state];
int cnt = 0, maxx = (shangxian ? shu[len] : 9);
for (int i = 0; i <= maxx; i++)
{
int tz = state;
if (state != 2 && i != 1)//不进行记录
tz = 0;
if (state != 2&& i == 1 )//标记当前位是1
tz = 1;
if (state== 1 && i == 3)//出现13，标记state=2，只要寻找出后面的数字组合;
tz = 2;
cnt += dfs(len - 1, (mod * 10 + i) % 13, tz, shangxian && i == maxx);
}
if (!shangxian)
dp[len][mod][state] = cnt;
return cnt;
}

int main()
{
while (~scanf("%d", &n))
{
memset(shu, 0, sizeof(shu));
memset(dp, 0, sizeof(dp));
int k = 0;
while (n)
{
shu[++k] = n % 10;
n /= 10;
}
printf("%d\n", dfs(k, 0, 0, 1));
}
return 0;
}