POJ 2545+2591+2247+1338简单水题

Posted 2020-09-18 mthoutai

【题意简述】：就是有这种一个序列。就拿当p1 = 2,p2 = 3, p3 = 5,来举例。由这三个数为基准组成的序列是:

2,3,4,5,6,8,9,10,12……如今给你这个序列数组的下标，让你求得这个数组中，这个下标里的数是多少。

【分析】：2,3,4,5,6,8,9,10,12……这个序列式由2,3,5这三个数生成的，详细怎样生成，就是：

详见代码：

这里以POJ2545为例：

//弄清当中的逻辑关系，能够从最简单的2,3,5试着做起！ 
#include<iostream>
#include<algorithm>
using namespace std;

int main()
{
	long long p1,p2,p3;
	int n;
	long long H[100005];
	int a1,a2,a3;
	
	while(cin>>p1>>p2>>p3>>n)
	{
		H[1] = 1;
		a1=a2=a3=1;
		for(int i = 2;i<=n+1;i++)
		{
			H[i] = min(p1*H[a1],min(p2*H[a2],p3*H[a3]));
			if(H[i]==p1*H[a1])  a1++;
			if(H[i]==p2*H[a2])  a2++;
			if(H[i]==p3*H[a3])  a3++;
		}
		cout<<H[n+1]<<endl;
	}
	return 0;
}

另外三道题与这个相差无几，都是这个的变形

仅仅只是2247 注意输出的格式！

//  C++ 代码 题目本身非常easy。注意格式输出。！
 
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

int main()
{
	int a1,a2,a3,a4;
	int n;
	long long humble[6000];
	humble[1]=1;
	a1=1;
	a2=1;
	a3=1;
	a4=1;
	for(int i=2;i<=5842;i++)
	{
		humble[i] = min(2*humble[a1],min(3*humble[a2],min(5*humble[a3],7*humble[a4])));
		if(humble[i]==2*humble[a1])
			a1++;
		if(humble[i]==3*humble[a2])
			a2++;
		if(humble[i]==5*humble[a3])
			a3++;
		if(humble[i]==7*humble[a4])
			a4++;	
	}
	//string b;
	while(1)
	{
		cin>>n;
		if(n==0)
			break;
		if(n%10==1)
		{
			if(n%100==11)
				cout<<"The "<<n<<"th"<<" humble number is "<<humble[n]<<"."<<endl;
			else
				cout<<"The "<<n<<"st"<<" humble number is "<<humble[n]<<"."<<endl;
		}
			
		if(n%10==2)
		{
			if(n%100==12)
				cout<<"The "<<n<<"th"<<" humble number is "<<humble[n]<<"."<<endl;
			else
				cout<<"The "<<n<<"nd"<<" humble number is "<<humble[n]<<"."<<endl;
		}
			
		if(n%10==3)
		{
			if(n%100==13)
				cout<<"The "<<n<<"th"<<" humble number is "<<humble[n]<<"."<<endl;
			else
				cout<<"The "<<n<<"rd"<<" humble number is "<<humble[n]<<"."<<endl;
		}
			
		if(n%10>3||n%10==0)
			cout<<"The "<<n<<"th"<<" humble number is "<<humble[n]<<"."<<endl;
			
	}
}

POJ  2591

// 39392K  141Ms
// 打表过得 =_= 
#include<iostream>
#include<algorithm>
using namespace std;
int S[10000001];

int main()
{
	int a,b,n;
	S[1] = 1;
	a = b = 1;
	for(int i = 2;i<=10000000;i++)
	{
		S[i] = min(2*S[a]+1,3*S[b]+1);
		if(S[i] == 2*S[a]+1) a++;
		if(S[i] == 3*S[b]+1) b++;
	}
	while(cin>>n)
	{
		cout<<S[n]<<endl;
	}
	return 0;
}