Humble Numbers

打的暴力，样例都要过好久，小脑直接萎缩© ©
Vjudge

• A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers.
  Write a program to find and print the nth element in this sequence
  Input
  The input consists of one or more test cases. Each test case consists of one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for n.
  Output
  For each test case, print one line saying "The nth humble number is number.". Depending on the value of n, the correct suffix "st", "nd", "rd", or "th" for the ordinal number nth has to be used like it is shown in the sample output.

• Sample

  Input	
  1
  2
  3
  4
  11
  12
  13
  21
  22
  23
  100
  1000
  5842
  

  Output
  The 1st humble number is 1.
  The 2nd humble number is 2.
  The 3rd humble number is 3.
  The 4th humble number is 4.
  The 11th humble number is 12.
  The 12th humble number is 14.
  The 13th humble number is 15.
  The 21st humble number is 28.
  The 22nd humble number is 30.
  The 23rd humble number is 32.
  The 100th humble number is 450.
  The 1000th humble number is 385875.
  The 5842nd humble number is 2000000000.
  

  脑残暴力

  点击查看代码

  bool check(int x)
  {
  //	cout<<x<<" ";
  	int tmp=-1;
  	while(x!=tmp&&x)
  	{	
  		tmp=x;
  		if(x%2==0)x/=2;
  		else if(x%3==0)x/=3;
  		else if(x%5==0)x/=5;
  		else if(x%7==0)x/=7;
  		else break;
  //		cout<<x<<endl;	
  	}
  //	cout<<tmp<<endl;
  	if(x&1&&x!=1)return 0;
  	else return 1;
  }
  signed main()
  {
  //	check(26);
  
  	while(scanf("%lld",&k)!=EOF&&k)
  	{
  		a[++cnt]=k;
  	}
  	int l=2;
  	b[1]=1;
  	for(int i=1;i<=cnt;i++)
  	{
  		if(bcnt<=a[i])
  //		else 
  		{
  			while(bcnt<a[i])
  			{
  				if(check(l))b[++bcnt]=l;
  //				cout<<bcnt<<" "<<b[bcnt]<<endl;
  				l++;
  			}
  //			cout<<b[a[i]]<<endl;
  		}
  		cout<<b[a[i]]<<endl;
  	}
  	return 0;
  }
  

  普通线性DP（也超时）

  点击查看代码

  
  const int N = 6000, mod = 1e9 + 7;
  const int inf = 0x3f3f3f3f;
  const double PI = acos(-1.0);
  
  int n;
  ll f[N];
  int a[4] = {2, 3, 5, 7};
  int main()
  {
  	memset(f, inf, sizeof f);
  	f[1] = 1;
  	for (int i = 2; i <= 5842; i ++ )
  	{
  		for (int j = 1; j <= i; j ++ )
  		{
  			for (int k = 0; k < 4; k ++ )
  			{
  				if (a[k] * f[j] > f[i - 1])
  				{
  					f[i] = min(f[i],a[k] * f[j]);
  				}
  			}
  		}
  	}
  	while (~scanf("%d", &n))
  	{
  		if (n == 0)  break;
  		if (n % 10 == 1 && n % 100 != 11) printf("The %dst humble number is ", n);
  		else if (n % 10 == 2 && n % 100 != 12) printf("The %dnd humble number is ", n);
  		else if (n % 10 == 3 && n % 100 != 13) printf("The %drd humble number is ", n);
  		else printf("The %dth humble number is ", n);
  		printf("%d.\n", f[n]);
  	}
  	return 0;
  }
  
  

  时间复杂度为k*n的DP 这里有一些坑，每一个数输出不一样(这是在考我英语？)
  点击查看代码

  void solve()
  {
  	dp[1]=a=b=c=d=1;
  	for(int i=2;i<=5842;i++)
  	{
  		dp[i]=min({dp[a]*2,dp[b]*3,dp[c]*5,dp[d]*7});
  		if(dp[i]==dp[a]*2)a++;
  		if(dp[i]==dp[b]*3)b++;
  		if(dp[i]==dp[c]*5)c++;
  		if(dp[i]==dp[d]*7)d++;
  	}
  }
  signed main()
  {
  	solve();
  	while(scanf("%lld",&n)!=EOF&&n)
  	{	
  		if (n == 0) break;
  		if (n % 10 == 1 && n % 100 != 11) printf("The %dst humble number is ", n);
  		else if (n % 10 == 2 && n % 100 != 12) printf("The %dnd humble number is ", n);
  		else if (n % 10 == 3 && n % 100 != 13) printf("The %drd humble number is ", n);
  		else printf("The %dth humble number is ", n);
  		printf("%lld.\n", dp[n]);
  //		cout<<dp[x]<<endl;
  	}
  	return 0;
  }