BestCoder Round #70 总结

Jam's math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 173    Accepted Submission(s): 88

Problem Description

Jam has a math problem. He just learned factorization.
He is trying to factorize  ax2+bx+c  into the form of  pqx2+(qk+mp)x+km=(px+k)(qx+m) .
He could only solve the problem in which p,q,m,k are positive numbers.
Please help him determine whether the expression could be factorized with p,q,m,k being postive.

 

Input

The first line is a number  T , means there are  T(1≤T≤100)  cases 

Each case has one line,the line has  3  numbers  a,b,c(1≤a,b,c≤100000000)

 

Output

You should output the "YES" or "NO".

 

Sample Input

   
   
   
   

    
    
    
    2
1 6 5
1 6 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    YES
NO


    
    
    
    
 
     
 
      Hint 
     
 The first case turn $x^2+6*x+5$ into $(x+1)(x+5)$ 
    
    
    
    
     
   
   
   
   

 

Source

BestCoder Round #70

 

题意：

Jam有道数学题想向你请教一下，他刚刚学会因式分解比如说,x^2+6x+5=(x+1)(x+5)x​2​​+6x+5=(x+1)(x+5)
就好像形如 ax^2+bx+cax​2​​+bx+c => pqx^2+(qk+mp)x+km=(px+k)(qx+m)pqx​2​​+(qk+mp)x+km=(px+k)(qx+m)
但是他很蠢，他只会做$p,q,m,k$为正整数的题目
请你帮助他，问可不可以分解

官方题解：第一道题比较简单，可以说是简单的模拟题，我们考虑到$a,b,c$都是10^{9}10​9​​的，所以我们决定要把时间复杂度降下来，对于每一个数，因为考虑到都是正数，所以我们处理起来就方便很多，打个比方$32=2\ast 16$，那么枚举到$2$的时候就可以得出$16$，这样子的话时间就变为O(\sqrt{a}\sqrt{b})O(√​a​​​√​b​​​)，轻松解决这道题

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        int ok=0;
        for(int i=1;i<=a/2+1;i++)
        {
            if(a%i==0)
            {
                for(int j=1;j<=c/2+1;j++)
                {
                    if(c%j==0)
                    {
                        int x=a/i;
                        int y=c/j;
                        if(x*j+y*i==b)
                        {
                            ok=1;
                            break;
                        }
                    }
                }
            }
            if(ok)
                break;
        }
        if(ok)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}
Jam's balance
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 154    Accepted Submission(s): 67



   
   
   
   

    
    
    
    Problem Description
   
   
   
   
   
   
   
   

    
    
    
    Jim has a balance and N weights. 
    
    
    
    
    
    
    
     (1≤N≤20) 
    
    
    
    
The balance can only tell whether things on different side are the same weight.
    
    
    
    
Weights can be put on left side or right side arbitrarily.
    
    
    
    
Please tell whether the balance can measure an object of weight M.
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   

   
   
   
   

    
    
    
    Input
   
   
   
   
   
   
   
   

    
    
    
    The first line is a integer 
    
    
    
    
    
    
    
     T(1≤T≤5) , means T test cases.
    
    
    
    
For each test case :
    
    
    
    
The first line is 
    
    
    
    
    
    
    
     N , means the number of weights.
    
    
    
    
The second line are 
    
    
    
    
    
    
    
     N  number, i'th number 
    
    
    
    
    
    
    
     wi(1≤wi≤100)  means the i'th weight's weight is 
    
    
    
    
    
    
    
     wi .
    
    
    
    
The third line is a number 
    
    
    
    
    
    
    
     M . 
    
    
    
    
    
    
    
     M  is the weight of the object being measured.
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   

   
   
   
   

    
    
    
    Output
   
   
   
   
   
   
   
   

    
    
    
    You should output the "YES"or"NO".
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   

   
   
   
   

    
    
    
    Sample Input
   
   
   
   
   
   
   
   

    
    
    
    
     
     
     
     

      
      
      
      1
2
1 4
3
2
4
5
     
     
     
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   

   
   
   
   

    
    
    
    Sample Output
   
   
   
   
   
   
   
   

    
    
    
    
     
     
     
     

      
      
      
      NO
YES
YES


      
      
      
      
 
       
 
        Hint 
       
 For the Case 1:Put the 4 weight alone For the Case 2:Put the 4 weight and 1 weight on both side 
      
      
      
      
       
     
     
     
     
   
   
   
   
题意：
Jam有$N$个砝码和一个没有游标的天平，现在给他$(1\le N\le 20)$个砝码，砝码可以放左边，也可以放右边，问可不可以测出所问的重量, 问的个数为$(1\le M\le 100)$个.
官方题解：这道题可以放左边，可以放右边，$N=20$显然每种状态都枚举是不太现实的，因为每组砝码都可以变成很多种重量，当然也不排除有人乱搞过了这一题，其实这道题是一道贪心的思想，我们看到$w$不大，所以可以用$01$背包扫一次，当然这还是不够的，这只能放一边，考虑到可以放另一边，就是可以有减的关系，所以反着再背包一遍，注意要判断边界。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxm=2e3+10;
int dp[maxm];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        int n,a;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a);
            for(int j=0;j<=2000;j++)
            {
                if(dp[j]&&j>a)
                {
                    dp[j-a]=1;
                }
            }
            for(int j=2000;j>=0;j--)
            {
                if(dp[j]==1)
                {
                    dp[j+a]=1;
                }
            }
        }
        int q;
        scanf("%d",&q);
        for(int i=0;i<q;i++)
        {
            int x;
            scanf("%d",&x);
            if(x<0||x>2000||!dp[x])
                printf("NO\n");
            else
                printf("YES\n");
        }
    }
    return 0;
}