hdu 3652 B-number （数位dp+记忆化）

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2143    Accepted Submission(s): 1152

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

 

Author

wqb0039

 

Source

2010 Asia Regional Chengdu Site —— Online Contest

 

Recommend

lcy

 

#include 
#include 
#include 
#define LL long long
using namespace std;

const int MAXN = 15;
const int MOD = 13;
int dp[MAXN][MAXN][MAXN][2],digit[MAXN],cnt;

int solve(int pos, int pre, int mod, bool flag, bool limit){
    if (!pos) return !mod && flag;
    if (!limit && dp[pos][pre][mod][flag]!=-1)
        return dp[pos][pre][mod][flag];

    int val_sz = limit ? digit[pos] : 9;
    int sum = 0;
    for (int i = 0; i <= val_sz; ++i){
        int lim = limit && (i == digit[pos]);
        if (pre == 1 && i == 3)
            sum += solve(pos - 1, i, (mod * 10 + i) % MOD, 1, lim);
        else
            sum += solve(pos - 1, i, (mod * 10 + i) % MOD, flag, lim);
    }
    if (!limit) dp[pos][pre][mod][flag] = sum;
    return sum;
}

void init_digits(int x){
    memset(digit, 0, sizeof(digit));
    cnt = 0;
    while (x){
        digit[++cnt] = x % 10;
        x /= 10;
    }
}

void check(int x){
    int sum = 0;
    for (int i = 1; i <= x; ++i){
        if  (i % MOD) continue;
        init_digits(i);
        int ok = 0;
        for (int j = cnt; j; --j)
            if (j < cnt && digit[j] == 3 && digit[j+1] == 1){
                ok = 1; break;
            }
        sum += ok;
    }
    printf("%d\n", sum);
}

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

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