I Count Two Three

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3582 Accepted Submission(s): 1094

Problem Description

I will show you the most popular board game in the Shanghai Ingress Resistance Team.
It all started several months ago.
We found out the home address of the enlightened agent Icount2three and decided to draw him out.
Millions of missiles were detonated, but some of them failed.

After the event, we analysed the laws of failed attacks.
It's interesting that the

i-th attacks failed if and only if

i can be rewritten as the form of

2a3b5c7d which

a,b,c,d are non-negative integers.

At recent dinner parties, we call the integers with the form

2a3b5c7d "I Count Two Three Numbers".
A related board game with a given positive integer

n from one agent, asks all participants the smallest "I Count Two Three Number" no smaller than

Input

The first line of input contains an integer

t (1≤t≤500000), the number of test cases.

t test cases follow. Each test case provides one integer

n (1≤n≤109).

Output

For each test case, output one line with only one integer corresponding to the shortest "I Count Two Three Number" no smaller than

Sample Input

  10 1 11 13 123 1234 12345 123456 1234567 12345678 123456789
 

Sample Output

  1 12 14 125 1250 12348 123480 1234800 12348000 123480000
 

思路：暴力解题利用lower_bound求解

#include<stdio.h>
#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm>  
#include<string>   
#include<iomanip>  
#include<cmath>
#include<conio.h>
#include<set>
#define MAX 1000000005
typedef  long long ll; 
using namespace std; 


int main()
{
	ll n;
	set <ll> p;
	set<ll>::iterator itlow; 
	ll t1,t2,t3,t4;
	for( t1 =1;t1 < MAX;t1*=2)
	{	
		p.insert(t1);
		for(t2=t1;t2<MAX;t2*=3)
		{
			p.insert(t2);
			for(t3=t2;t3<MAX;t3*=5)
			{
				p.insert(t3);
				for(t4=t3;t4<MAX;t4*=7)
				{
					p.insert(t4);
			 	} 
			} 
		}  
	} 
	
	ll t;
	scanf("%lld",&t);
	while(t--)
	{
		scanf("%lld",&n);
		itlow = p.lower_bound(n);
		printf("%lld\n",*itlow);	
	} 
	return 0;
}

Cure

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7400 Accepted Submission(s): 1099

Problem Description

Given an integer

n, we only want to know the sum of

1/k2 where

k from

1 to

Input

There are multiple cases.
For each test case, there is a single line, containing a single positive integer

n.
The input file is at most 1M.

Output

The required sum, rounded to the fifth digits after the decimal point.

Sample Input

  1 2 4 8 15
 

Sample Output

  1.00000 1.25000 1.42361 1.52742 1.58044
 

思路暴力打表，趋近无穷值为 PI*PI/6

#include<stdio.h>
#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm>  
#include<string>   
#include<iomanip>  
#include<cmath>
#include<conio.h>
#include<set>
#define PI 3.1415926535897932384626433

typedef  long long ll; 
using namespace std; 

double a[200005];

int main()
{
	memset(a,0,sizeof(a));
	for(unsigned long long i=1;i<200005;i++)
	{
	    double t = 1.0/(i*i);
		a[i] = a[i-1] + t;
	}
	
	unsigned long long  n;
	while(scanf("%llu",&n) == 1)
	{
		if(n>200000)
			printf("%.5lf\n",PI*PI/6);
		else
			printf("%.5lf\n",a[n]);
	}
	return 0;
}

Balanced Game

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2126 Accepted Submission(s): 1448

Problem Description

Rock-paper-scissors is a zero-sum hand game usually played between two people, in which each player simultaneously forms one of three shapes with an outstretched hand. These shapes are "rock", "paper", and "scissors". The game has only three possible outcomes other than a tie: a player who decides to play rock will beat another player who has chosen scissors ("rock crushes scissors") but will lose to one who has played paper ("paper covers rock"); a play of paper will lose to a play of scissors ("scissors cut paper"). If both players choose the same shape, the game is tied and is usually immediately replayed to break the tie.

Recently, there is a upgraded edition of this game: rock-paper-scissors-Spock-lizard, in which there are totally five shapes. The rule is simple: scissors cuts paper; paper covers rock; rock crushes lizard; lizard poisons Spock; Spock smashes scissors; scissors decapitates lizard; lizard eats paper; paper disproves Spock; Spock vaporizes rock; and as it always has, rock crushes scissors.

Both rock-paper-scissors and rock-paper-scissors-Spock-lizard are balanced games. Because there does not exist a strategy which is better than another. In other words, if one chooses shapes randomly, the possibility he or she wins is exactly

50% no matter how the other one plays (if there is a tie, repeat this game until someone wins). Given an integer

N, representing the count of shapes in a game. You need to find out if there exist a rule to make this game balanced.

Input

The first line of input contains an integer t, the number of test cases. t test cases follow.
For each test case, there is only one line with an integer

N (2≤N≤1000), as described above.

Here is the sample explanation.

In the first case, donate two shapes as A and B. There are only two kind of rules: A defeats B, or B defeats A. Obviously, in both situation, one shapes is better than another. Consequently, this game is not balanced.

In the second case, donate two shapes as A, B and C. If A defeats B, B defeats C, and C defeats A, this game is balanced. This is also the same as rock-paper-scissors.

In the third case, it is easy to set a rule according to that of rock-paper-scissors-Spock-lizard.

Output

For each test cases, output "Balanced" if there exist a rule to make the game balanced, otherwise output "Bad".

Sample Input

  3 2 3 5
 

Sample Output

  Bad Balanced Balanced
 

判断数的奇偶性，奇数为Bad 偶数为Balanced