首页 > 代码库 > hdu 3652 B-number
hdu 3652 B-number
B-number
http://acm.hdu.edu.cn/showproblem.php?pid=3652
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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
Source
2010 Asia Regional Chengdu Site —— Online Contest
Recommend
lcy | We have carefully selected several similar problems for you: 3651 3655 3654 3653 3659
数位DP
#include<cstdio>#include<cstring>using namespace std;int a[20],pos,dp[20][3][13];int dfs(int dep,int type,int mod,bool lim){ if(!dep) return type==2 && !mod; if(!lim && dp[dep][type][mod]!=-1) return dp[dep][type][mod]; int ans=0,up= lim ? a[dep] : 9; for(int i=0;i<=up;i++) { if(type==2 || (type==1 && i==3)) ans+=dfs(dep-1,2,(mod*10+i)%13,lim && i==up); else ans+=dfs(dep-1,i==1 ? 1 : 0,(mod*10+i)%13,lim && i==up); } if(!lim) dp[dep][type][mod]=ans; return ans;}int solve(int n){ while(n) { a[++pos]=n%10; n/=10; } return dfs(pos,0,0,1);}int main(){ int n; memset(dp,-1,sizeof(dp)); while(scanf("%d",&n)!=EOF) { printf("%d\n",solve(n)); pos=0; }}
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。