hdu3652(数位dp,记忆化搜索)

题目描述:

B-number
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12235    Accepted Submission(s): 7320


Problem Description
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
13
100
200
1000
 

Sample Output
1
1
2
2

思路:记忆化搜索

#include<iostream>
#include<string.h>
#include<cmath>
#include<set>
#include<map>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
using namespace std;
typedef long long ll;
inline ll read() {
    ll sum = 0, f = 1;
    char p = getchar();
    for (; !isdigit(p); p = getchar()) if (p == '-')f = -1;
    for (; isdigit(p); p = getchar())  sum = sum * 10 + p - 48;
    return sum * f;
}
const int maxn = 50;
int n,dig[maxn],len,res;
int dp[maxn][13][3];
int dfs(int len, int ismax,int last,int has13) {
    int ans = 0, maxx,has,pre,lastx;
    if (!len)return last == 0 && has13==2;
    if (!ismax&&dp[len][last][has13] != -1)return dp[len][last][has13];
    maxx = ismax ? dig[len] : 9;
    for (int i = 0; i <= maxx; i++) {
        lastx = (last * 10 + i) % 13;
        if (has13 == 2)has = 2;
        else if (has13 == 1 && i == 3)has = 2;
        else if (i == 1)has = 1;
        else has = 0;
        ans += dfs(len - 1, ismax && i == maxx, lastx,has);
    }
    if (!ismax)dp[len][last][has13] = ans;
    return ans;
}
int main() {
    //freopen("test.txt", "r", stdin);
    while (~scanf("%d", &n)) {
        res = len = 0;
        memset(dp, -1, sizeof(dp));
        while (n){
            dig[++len] = n % 10;
            n /= 10;
        }dig[len + 1] = 0;
        printf("%d\n", dfs(len, 1, 0, 0));
    }
    return 0;
}