HDU 4028 The time of a day (数论，离散DP）

The time of a day

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 742    Accepted Submission(s): 327

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 <= 2⁶³-1)

 

 

Output

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

 

 

Sample Input

3 5 5 10 1 10 128

 

 

Sample Output

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

 

 

Source

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

 

 

Recommend

lcy

 

 

离散DP。

就是求1-n中任意选取若干个数的LCM

 

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<map>
using namespace std;

map<long long,long long>dp[45];

long long gcd(long long a,long long b)
{
    if(b==0)return a;
    return gcd(b,a%b);
}
long long lcm(long long a,long long b)
{
    return a*b/gcd(a,b);
}

int main()
{
    dp[1][1]=1;
    map<long long,long long>::iterator it;
    for(int i=2;i<=40;i++)
    {
        dp[i]=dp[i-1];
        dp[i][i]++;
        for(it=dp[i-1].begin();it!=dp[i-1].end();it++)
        {
            dp[i][lcm(it->first,i)]+=it->second;
        }
    }
    int T;
    int iCase=0;
    scanf("%d",&T);
    int n;
    long long m;
    while(T--)
    {
        iCase++;
        scanf("%d%I64d",&n,&m);
        long long ans=0;
        for(it=dp[n].begin();it!=dp[n].end();it++)
          if(it->first>=m)
            ans+=it->second;
        printf("Case #%d: %I64d\n",iCase,ans);
    }
    return 0;
}