CF55D:Beautiful numbers(数位dp + 数论)
2017-02-09 22:14
295 查看
reference:http://blog.csdn.net/lvshubao1314/article/details/43907225
D. Beautiful numbers
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue
with this and just count the quantity of beautiful numbers in given ranges.
Input
The first line of the input contains the number of cases t (1 ≤ t ≤ 10).
Each of the next t lines contains two natural numbers li and ri(1 ≤ li ≤ ri ≤ 9 ·1018).
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also
you may use %I64d).
Output
Output should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals
(from li to ri,
inclusively).
Examples
input
output
input
output
题意:求区间内有多少个数,其本身能被各个位上的数字整除。
思路:略。
D. Beautiful numbers
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue
with this and just count the quantity of beautiful numbers in given ranges.
Input
The first line of the input contains the number of cases t (1 ≤ t ≤ 10).
Each of the next t lines contains two natural numbers li and ri(1 ≤ li ≤ ri ≤ 9 ·1018).
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also
you may use %I64d).
Output
Output should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals
(from li to ri,
inclusively).
Examples
input
1 1 9
output
9
input
1 12 15
output
2
题意:求区间内有多少个数,其本身能被各个位上的数字整除。
思路:略。
# include <stdio.h> # include <string.h> # define LL long long # define MOD 2520 LL a[21], index[2521], dp[21][MOD][50]; void init() { int num = 0; for(int i=1; i<=MOD; ++i) if(!(MOD%i)) index[i] = num++;//最小公倍数离散化,优化内存。 } int gcd(int a, int b) { if(b==0) return a; return gcd(b, a%b); } int lcm(int a, int b) { return a/gcd(a, b)*b; } LL dfs(int pos, int sum, int prelcm, bool limit) { if(pos == -1) return sum%prelcm==0; if(!limit && dp[pos][sum][index[prelcm]] != -1) return dp[pos][sum][index[prelcm]]; int up = limit?a[pos]:9; LL ans = 0; for(int i=0; i<=up; ++i) { int newsum = (sum*10+i)%MOD;//2520为1~9的最小公倍数,数值对2520取余,如果该数(比2520大)能被其各位数整除,则它一定能被2520整除。 int newlcm = prelcm; if(i) newlcm = lcm(newlcm, i); ans += dfs(pos-1, newsum, newlcm, limit&&(i==a[pos])); } if(!limit) dp[pos][sum][index[prelcm]] = ans; return ans; } LL solve(LL num) { int len = 0; while(num) { a[len++] = num%10; num /= 10; } return dfs(len-1, 0, 1, true); } int main() { LL n, m; int t; init(); memset(dp, -1, sizeof(dp)); scanf("%d",&t); while(t--) { scanf("%I64d%I64d",&n,&m); printf("%I64d\n",solve(m)-solve(n-1)); } return 0; }
相关文章推荐
- 第六十六课、c++中的类型识别
- Linux系统中的文件权限
- 深入研究容器
- C++标准库:<algorithm>next_permutation() HDU - 1027
- java多线程之继承Thread类创建线程类
- 初识NuGet - 概念, 安装和使用
- [Boolan]第二周学习笔记——rico风
- 设计模式字典
- 码农日记-1
- Android权限Uri.parse的几种用法
- Boto3 操作AWS的SQS
- 学习SQLite基础笔记
- Android官方资料--Configuration Examples
- XML序列之System.Xml.Serialization
- java 线程通信
- numpy 矩阵数据共享知识小结
- 网站前端_EasyUI.基础入门.0007.使用EasyUI Tabs组件的最佳姿势?
- canvas粒子时钟
- ASP.NET MVC (六-1) Razor视图引擎
- 关于【error C3646: 未知重写说明符】的若干种可能性