hdu 4028 The time of a day 求前n个数的lcm>=m的子集个数

本文链接：https://blog.youkuaiyun.com/kongming_acm/article/details/6771670

本文深入解析了银币岛居民如何通过指针钟表精确定义日长，及其优化日长设置以满足特定需求的数学原理与方法论。详细介绍了计算最优日长的方法及实现步骤。

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Problem Description

There are no days and nights on byte island, so the residents here can hardly determine the length of a single day. Fortunately, they have invented a clock with several pointers. They have N pointers which can move round the clock. Every pointer ticks once per second, and the i-th pointer move to the starting position after i times of ticks. The wise of the byte island decide to define a day as the time interval between the initial time and the first time when all the pointers moves to the position exactly the same as the initial time.
The wise of the island decide to choose some of the N pointers to make the length of the day greater or equal to M. They want to know how many different ways there are to make it possible.

Input

There are a lot of test cases. The first line of input contains exactly one integer, indicating the number of test cases.
For each test cases, there are only one line contains two integers N and M, indicating the number of pointers and the lower bound for seconds of a day M. (1 <= N <= 40, 1 <= M <= 2 ⁶³-1)

Output

For each test case, output a single integer denoting the number of ways.

Sample Input

Sample Output

  
  
   
   Case #1: 22
Case #2: 1023
Case #3: 586

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
long long gcd(long long a,long long b)
{
    return b==0?a:gcd(b,a%b);
}
long long lcm(long long a,long long b)
{
    return a/gcd(a,b)*b;
}
map<long long ,long long > dp[50];
//dp[i][j]表示前i个数，最小公倍数为j的个数
//离散DP
void DP()
{
    dp[1][1]=1;
    for(int i=2;i<=40;i++)
    {
        dp[i]=dp[i-1];
        dp[i][i]++;
        map<long long ,long long >::iterator j;
        for(j=dp[i-1].begin();j!=dp[i-1].end();++j)
        {
            long long lc=lcm(i,j->first);
            dp[i][lc]+=j->second;
        }
    }
}
int main()
{
    DP();
    int ci,pl=1;scanf("%d",&ci);
    while(ci--)
    {
        int n;long long m;
        scanf("%d%I64d",&n,&m);
        long long ans=0;
        map<long long ,long long >::iterator j;
        for(j=dp[n].begin();j!=dp[n].end();++j)
        {
            if(j->first>=m) ans+=j->second;
        }
        printf("Case #%d: %I64d\n",pl++,ans);
    }
    return 0;
}