hdu3652 B-number 数位dp

题目：

http://acm.hdu.edu.cn/showproblem.php?pid=3652

题意：

给出一个数n，求从1到n范围内的数字含有子串13且能被13整除的个数

思路：

数位dp，再加一维取模的状态，我对动态规划好迷啊。。。

#include <bits/stdc++.h>using namespace std;const int N = 50 + 10;int dp[N][13][2][2];int dig[N];int dfs(int pos, int mod, bool pre, bool status, bool limit){    if(pos < 1) return mod == 0 && status;    if(! limit && dp[pos][mod][pre][status] != -1) return dp[pos][mod][pre][status];    int en = limit ? dig[pos] : 9;    int ans = 0;    for(int i = 0; i <= en; i++)        ans += dfs(pos-1, (mod*10+i) % 13, i == 1, status || (pre == 1 && i == 3), limit && i == en);    if(! limit) dp[pos][mod][pre][status] = ans;    return ans;}int work(int n){    int tot = 0;    while(n) dig[++tot] = n % 10, n /= 10;    memset(dp, -1, sizeof dp);    return dfs(tot, 0, 0, 0, 1);}int main(){    int n;    while(~ scanf("%d", &n))        printf("%d\n", work(n));    return 0;}

#include <bits/stdc++.h>using namespace std;const int N = 50 + 10;int dp[N][13][3];int dig[N];int dfs(int pos, int mod, int status, bool limit){    if(pos < 1) return mod == 0 && status == 2;    if(! limit && dp[pos][mod][status] != -1) return dp[pos][mod][status];    int en = limit ? dig[pos] : 9;    int ans = 0;    for(int i = 0; i <= en; i++)    {        int t = status;        if(t != 2)        {            if(i == 1) t = 1;            else if(t == 1 && i == 3) t = 2;            else t = 0;        }        ans += dfs(pos-1, (mod*10 + i) % 13, t, limit && i == en);    }    if(!limit) dp[pos][mod][status] = ans;    return ans;}int work(int n){    int tot = 0;    while(n) dig[++tot] = n % 10, n /= 10;    memset(dp, -1, sizeof dp);    return dfs(tot, 0, 0, 1);}int main(){    int n;    while(~ scanf("%d", &n))        printf("%d\n", work(n));    return 0;}