• CF912A Tricky Alchemy


    题意翻译

    Grisha有一些水晶,可以用这些水晶造出一些水晶球。现在他有A个黄水晶,B个蓝水晶。现在他要造出x个黄水晶球,y个绿水晶球,z个蓝水晶球。请问他还额外需要几个水晶(不需要就输出0)才能完成任务。

    2个黄水晶可造出1个黄水晶球

    1个黄水晶加上1个蓝水晶可造出1个绿水晶球

    3个蓝水晶可造出1个蓝水晶球

    题目描述

    During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 20182018 , 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 AA yellow and BB blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.

    输入输出格式

    输入格式:

     

    The first line features two integers AA and BB ( 0<=A,B<=10^{9}0<=A,B<=109 ), denoting the number of yellow and blue crystals respectively at Grisha's disposal.

    The next line contains three integers xx , yy and zz ( 0<=x,y,z<=10^{9}0<=x,y,z<=109 ) — the respective amounts of yellow, green and blue balls to be obtained.

     

    输出格式:

     

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

     

    输入输出样例

    输入样例#1: 复制
    4 3
    2 1 1
    
    输出样例#1: 复制
    2
    
    输入样例#2: 复制
    3 9
    1 1 3
    
    输出样例#2: 复制
    1
    
    输入样例#3: 复制
    12345678 87654321
    43043751 1000000000 53798715
    
    输出样例#3: 复制
    2147483648
    

    说明

    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.

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    long long a,b,x,y,z;
    long long ans,sum1,sum2;
    int main(){
        cin>>a>>b;
        cin>>x>>y>>z;
        sum1+=x*2;sum1+=y;
        sum2+=y;sum2+=z*3;
        ans+=(sum1-a)>0?sum1-a:0;
        ans+=(sum2-b)>0?sum2-b:0;
        cout<<ans;
    }
    细雨斜风作晓寒。淡烟疏柳媚晴滩。入淮清洛渐漫漫。 雪沫乳花浮午盏,蓼茸蒿笋试春盘。人间有味是清欢。
  • 相关阅读:
    PostgreSQL-14-异常值处理
    Python-5-字符串方法
    Python-6-字典-函数dict,字典的基本操作及将字符串设置功能用于字典
    STP-6-快速生成树协议-新端口角色,状态和类型以及新链路类型
    PostgreSQL-13-缺失值处理
    IP服务-6-SNMP
    IP服务-7-系统日志
    Python-4-设置字符串的格式字符串
    IP服务-5-网络时间协议
    Python -3-列表和元组
  • 原文地址:https://www.cnblogs.com/cangT-Tlan/p/8443604.html
Copyright © 2020-2023  润新知