HDU 3652 B-number (数位DP)

题意：给定一个数 n，求1-n之间有多少个包含13，并且是13的倍数的数。

析：数位DP，dp[i][j][k]，表示前 i 位，模13余数为 j，k = 0，表示不含 13并且前一位不是1，k = 1，表示不含13但前一位是1，k = 2，

表示含13，那么剩下的就简单了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[15][15][3], a[15];

int dfs(int pos, int last, int val, bool ok){
    if(!pos)  return last == 0 && val == 2;
    int &ans = dp[pos][last][val];
    if(!ok && ans >= 0)  return ans;

    int res = 0, n = ok ? a[pos] : 9;
    for(int i = 0; i <= n; ++i){
        if(i == 3 && val == 1)  res += dfs(pos-1, (last*10+i)%13, 2, ok && i == n);
        else if(val == 2)  res += dfs(pos-1, (last*10+i)%13, 2, ok && i == n);
        else if(i == 1 && !val)  res += dfs(pos-1, (last*10+i)%13, 1, ok && i == n);
        else if(i == 1) res += dfs(pos-1, (last*10+i)%13, 1, ok && i == n);
        else  res += dfs(pos-1, (last*10+i)%13, 0, ok && i == n);
    }
    if(!ok)  ans = res;
    return res;
}

int solve(int n){
    int len = 0;
    while(n){
        a[++len] = n % 10;
        n /= 10;
    }
    return dfs(len, 0, 0, true);
}

int main(){
    memset(dp, -1, sizeof dp);
    while(scanf("%d", &n) == 1){
        printf("%d\n", solve(n));
    }
    return 0;
}