codeforces 912A Tricky Alchemy

最新推荐文章于 2020-05-01 22:21:24 发布

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最新推荐文章于 2020-05-01 22:21:24 发布

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分类专栏： ACM算法_水题文章标签： codeforces 水题

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本文介绍了一个基于魔法晶体制造黄色、绿色和蓝色球体的问题。通过数学计算确定额外所需的黄色和蓝色魔法晶体数量，以满足目标球体的制作需求。

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During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.

Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.

Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.

Input

The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.

The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.

Output

Print a single integer — the minimum number of crystals that Grisha should acquire in addition.

   Examples
 
     input 
    
      Copy
    
4 3
2 1 1

     output 
    
      Copy
    
2

     input 
    
      Copy
    
3 9
1 1 3

     output 
    
      Copy
    
1

     input 
    
      Copy
    
12345678 87654321
43043751 1000000000 53798715

     output 
    
      Copy
    
2147483648

Note

In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.

水题，注意超出了int，用lang lang

#include<bits/stdc++.h>  
using namespace std;  
typedef long long ll;  
const int inf = 0x3f3f3f3f;  
int main()   
{  
    // freopen("shuju.txt","r",stdin);
    ll a,b;
    ll x,y,z;

    cin>>a>>b;
    cin>>x>>y>>z;  

    ll ans=0;
    ll p=0,q=0;
    p+=2*x;
    p+=y;
    q+=y;
    q+=3*z;
    //printf("## %d %d\n",p,q);
    if(p>a)
        ans+=p-a;
    if(q>b)
        ans+=q-b;
    cout<<ans<<endl;
    return 0;  
}