【CUGBACM15级BC第31场 B】hdu 5179 beautiful number-优快云博客

本文介绍了一种使用数位DP算法解决特定数学问题的方法——寻找区间内的美丽数。美丽数是一类特殊的整数，其每位数字需满足特定条件。文章详细解释了问题背景，并提供了一个完整的C++实现案例。

beautiful number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 851 Accepted Submission(s): 553

Problem Description

Let

A=∑ni=1ai∗10n−i(1≤ai≤9)(

n is the number of

A's digits). We call

A as “beautiful number” if and only if

a[i ]≥a[i+1] when

1≤i<n and

a[i ] mod

a[ j ]=0 when

1≤i≤n,i<j≤n(Such as 931 is a "beautiful number" while 87 isn't).
Could you tell me the number of “beautiful number” in the interval

[L,R ](including L and R)?

Input

The fist line contains a single integer

T(about 100), indicating the number of cases.
Each test case begins with two integers

L,R(1≤L≤R≤109).

Output

For each case, output an integer means the number of “beautiful number”.

Sample Input

2
1 11
999999993 999999999

Sample Output

10
2

数位dp

#include <cstring>
#include <cstdio>
#include <string.h>
#include <iostream>
using namespace std;

int num[15];
int dp[15][15];
///mark==1没前导0 mark==0有前导0
int dfs(int pos, int pre, int over, int mark)
{
    if (pos < 0)
    {
        return 1;
    }
    if (dp[pos][pre] != -1 && !over)
    {
        return dp[pos][pre];
    }
    int last = over ? num[pos] : 9;
    int ans = 0;
    for (int i = 0; i <= last; i++)
    {
        if (mark == 0 || (pre >= i && i != 0 && pre % i == 0))
        {
            ans += dfs(pos - 1, i, over && i == last, mark || i);
        }
    }
    if (!over)
    {
        dp[pos][pre] = ans;
    }
    return ans;
}
int solve(int n)
{
    int len = 0;
    while (n)
    {
        num[len++] = n % 10;
        n /= 10;
    }
    return dfs(len - 1, 0, true, 0);

}
int main()
{
    int t;
    scanf("%d", &t);
    memset(dp, -1, sizeof(dp));
    while (t--)
    {
        int l, r;
        scanf("%d%d", &l, &r);
        printf("%d\n", solve(r) - solve(l - 1));

    }
}