HDU 3652 B-number 数位dp

B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6350 Accepted Submission(s): 3678

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

题目大意：求不大于n的数中有多少个数既有13，又能被13整除。

不难看出这是一道数位dp，如果去掉被13整除这个条件，就是很裸的数位dp入门题，记忆化搜索即可。
但如果加上13这个条件后还像原来三维转移肯定是有bug的，比如13这个数，它是满足条件的，但113显然不满足，不过如果像原来那样转移就会认为113也是满足条件的。
考虑再加一维，表示对13取余后的结果。
f[i][j][k][z] i表示位数 j表示前一位数 k表示取mod z表示是否包含13
每次推到下一位的时候重新算一下模数就好。
也算一道板吧。
update:10.5

#include<cstdio>#include<cstring>#include<algorithm>#define ms(x,y) memset(x,y,sizeof(x))using namespace std;const int N = 15;int a[N];int dp[N][N][2][N];int dfs(int pos,int pre,int mod,int statue,bool zero,bool limit){    if(pos<=0) return (mod==0)&&statue;    if(!limit&&!zero&&dp[pos][pre][statue][mod]!=-1) return dp[pos][pre][statue][mod];    int mx;if(limit) mx=a[pos];    else mx=9;    int ans=0;    for(int i=0;i<=mx;i++){        ans+=dfs(pos-1,i,(mod*10%13+i%13)%13,statue||(pre==1&&i==3),zero&&i==0,limit&&i==mx);        //mod * 10 我之前是怎么想的 pre*10啊     }    if(!limit) dp[pos][pre][statue][mod]=ans;    return ans;}int query(int x){    int cnt=0;    while(x){        a[++cnt]=x%10;        x/=10;    }    ms(dp,-1);    return dfs(cnt,0,0,0,1,1);}int main(){    int n;;    while(~scanf("%d",&n)){       printf("%d\n",query(n));    }    return 0;}