首页 > 代码库 > 数位DP [HDU 3652] B-number
数位DP [HDU 3652] B-number
B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2668 Accepted Submission(s): 1467
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
数位DP模板题、只不过好久好久没写数位DP了、都忘了
#include<iostream>#include<cstdio>#include<cstring>using namespace std;int bit[15];int dp[15][15][3];int dfs(int pos,int s1,int s2,bool limit) //s1代表余数，s2代表状态{ if(pos==-1) { return (s1==0 && s2==2); } if(!limit && dp[pos][s1][s2]!=-1) return dp[pos][s1][s2]; int ans=0; int end=limit?bit[pos]:9; for(int i=0;i<=end;i++) { int nows2=0; if(s2==0) { if(i==1) nows2=1; } else if(s2==1) { if(i==1) nows2=1; else if(i==3) nows2=2; } else if(s2==2) nows2=2; ans+=dfs(pos-1,(s1*10+i)%13,nows2,limit && i==end); } if(!limit) dp[pos][s1][s2]=ans; return ans;}int cal(int n){ int len=0; while(n) { bit[len++]=n%10; n/=10; } return dfs(len-1,0,0,1);}int main(){ int n; memset(dp,-1,sizeof(dp)); while(scanf("%d",&n)!=EOF) { printf("%d\n",cal(n)); } return 0;}
数位DP [HDU 3652] B-number
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。