Codeforces-632D Longest Subsequence
2016-03-24 12:37
323 查看
You are given array a with n elements
and the number m. Consider some subsequence of a and
the value of least common multiple (LCM) of its elements. Denote LCM as l. Find any longest subsequence of a with
the value l ≤ m.
A subsequence of a is an array we can get by erasing some elements of a.
It is allowed to erase zero or all elements.
The LCM of an empty array equals 1.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 106)
— the size of the array a and the parameter from the problem statement.
The second line contains n integers ai (1 ≤ ai ≤ 109)
— the elements of a.
Output
In the first line print two integers l and kmax (1 ≤ l ≤ m, 0 ≤ kmax ≤ n)
— the value of LCM and the number of elements in optimal subsequence.
In the second line print kmax integers
— the positions of the elements from the optimal subsequence in the ascending order.
Note that you can find and print any subsequence with the maximum length.
Examples
input
output
input
output
题目大意:
给你一个具有n个元素的序列a以及一个数m,找出能够使得所有子序列中最长的一个子序列,并使其满足该子序列的最小公倍数<=m。
解题思路:
这个题目在codeforces里面被归在了暴力(brute force),数学(math),数论(number theory)下,其实感觉就是一个思维题,不过是挺暴力的= =对于整个序列里的每一个元素,只要是大于m的就不需要考虑直接跳过,因为两个数a和b的最小公倍数lcm,必然是大于或者等于a或b的,那么对于所有小于等于m的元素,只需要把所有小于m的元素的因子找到,并且更新因子的个数,找到最终小于m的元素中,具有相同因子个数最多的元素,就是所找到的l,然后再对原数组遍历一遍,找到他的因子们的位置,就可以了。
and the number m. Consider some subsequence of a and
the value of least common multiple (LCM) of its elements. Denote LCM as l. Find any longest subsequence of a with
the value l ≤ m.
A subsequence of a is an array we can get by erasing some elements of a.
It is allowed to erase zero or all elements.
The LCM of an empty array equals 1.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 106)
— the size of the array a and the parameter from the problem statement.
The second line contains n integers ai (1 ≤ ai ≤ 109)
— the elements of a.
Output
In the first line print two integers l and kmax (1 ≤ l ≤ m, 0 ≤ kmax ≤ n)
— the value of LCM and the number of elements in optimal subsequence.
In the second line print kmax integers
— the positions of the elements from the optimal subsequence in the ascending order.
Note that you can find and print any subsequence with the maximum length.
Examples
input
7 8 6 2 9 2 7 2 3
output
6 5 1 2 4 6 7
input
6 4 2 2 2 3 3 3
output
2 3 1 2 3
题目大意:
给你一个具有n个元素的序列a以及一个数m,找出能够使得所有子序列中最长的一个子序列,并使其满足该子序列的最小公倍数<=m。
解题思路:
这个题目在codeforces里面被归在了暴力(brute force),数学(math),数论(number theory)下,其实感觉就是一个思维题,不过是挺暴力的= =对于整个序列里的每一个元素,只要是大于m的就不需要考虑直接跳过,因为两个数a和b的最小公倍数lcm,必然是大于或者等于a或b的,那么对于所有小于等于m的元素,只需要把所有小于m的元素的因子找到,并且更新因子的个数,找到最终小于m的元素中,具有相同因子个数最多的元素,就是所找到的l,然后再对原数组遍历一遍,找到他的因子们的位置,就可以了。
#include <cstdio> #include <cstring> using namespace std; const int maxn = 1e6 + 5; int num[maxn], a[maxn], sum[maxn]; int main() { int n, m; scanf("%d%d", &n, &m); memset(num, 0, sizeof(num)); memset(sum, 0, sizeof(sum)); for(int i = 0; i < n; ++i){ scanf("%d", &a[i]); if(a[i] <= m){ ++num[a[i]]; } } for(int i = 1; i <= m; ++i){ if(num[i]){ for(int j = i; j <= m; j += i){ sum[j] += num[i]; } } } int max_sum = 0, max_val = 0; for(int i = 1; i <= m; ++i){ if(sum[i] && max_sum < sum[i]){ max_val = i; max_sum = sum[i]; } } if(max_sum == 0){ printf("1 0\n\n"); }else{ int cas = 0; printf("%d %d\n", max_val, max_sum); for(int i = 0; i < n; ++i){ if(max_val % a[i] == 0){ if(cas){ printf(" "); } printf("%d", i + 1); ++cas; } } puts(""); } return 0; }如有错误,还请各位菊苣指出!
相关文章推荐
- 控件UI性能调优 -- SizeChanged不是万能的
- 酷狗音乐模仿还在继续之UI界面源码下载
- JS高程读书笔记-第一、二章-内附在线思维导图和quizlet卡片
- IOS中UITableViewCell使用详解
- 原生的强大DOM选择器querySelector - querySelector和querySelectorAll
- easyui解析$.parser.parse()的学习,使用地方
- iOS UIStepper 加减计数器
- UIStackView入门示例解析
- UE的注册流程
- easyui checkbox 初始化时设为已选的方法
- java的build
- hdu 5306 Gorgeous Sequence
- 一个无序整数数组中找到最长连续序列(Longest Consecutive Sequence)和两个元素使得相差最小
- UIImagePickerController控件自带按钮显示中文的办法
- pku1947rebuilding roads 树形DP
- cf#VK Cup 2015 - Qualification Round 2-C - Name Quest-贪心
- EasyUI两种动态添加tab Iframe页面的方法
- EasyUI两种动态添加tab Iframe页面的方法
- UIImageView的内容模式以及ImageNamed和imageWithContentsOfFile的区别
- iOS-UITextField中给placeholder动态设置颜色的四种方法