【CUGBACM15级BC第31场 B】hdu 5179 beautiful number

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 ] moda[ 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 integerT(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

21 11999999993 999999999

 

Sample Output

102

 

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0int 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));    }}