hdu 3652 B-number （数位dp+记忆化）

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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

131002001000

Sample Output

Author

wqb0039

Source

2010 Asia Regional Chengdu Site —— Online Contest

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#include <cstdio>#include <cstring>#include <algorithm>#define LL long longusing 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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