zoukankan      html  css  js  c++  java
  • CF#256(Div.2) A. Rewards

    A. Rewards
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Bizon the Champion is called the Champion for a reason.

    Bizon the Champion has recently got a present — a new glass cupboard with n shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has a1 first prize cups, a2 second prize cups and a3third prize cups. Besides, he has b1 first prize medals, b2 second prize medals and b3 third prize medals.

    Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules:

    • any shelf cannot contain both cups and medals at the same time;
    • no shelf can contain more than five cups;
    • no shelf can have more than ten medals.

    Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled.

    Input

    The first line contains integers a1a2 and a3 (0 ≤ a1, a2, a3 ≤ 100). The second line contains integers b1b2 and b3 (0 ≤ b1, b2, b3 ≤ 100). The third line contains integer n (1 ≤ n ≤ 100).

    The numbers in the lines are separated by single spaces.

    Output

    Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes).

    Sample test(s)
    input
    1 1 1
    1 1 1
    4
    
    output
    YES
    
    input
    1 1 3
    2 3 4
    2
    
    output
    YES
    
    input
    1 0 0
    1 0 0
    1
    
    output
    NO  
    
    
    
    
    题意就是Bizon the Champion这个人得了非常多奖,有a1个一等奖奖杯。a2个二等奖奖杯。a3个三等奖奖杯。b1张一等奖奖状。b2张二等奖奖状,b3张三等奖奖状。如今给你n个柜子。问你能否将这些奖杯和奖状放下,规则是奖杯和奖状不能放在同一个柜子里。一个柜子最多仅仅能放5个奖杯或10张奖状。

    解题思路:将现有的奖杯和奖状所须要的柜子书求出。假设小于n,则输出“YES”;否则输出“NO”。
    #include<stdio.h>
    int s[3],y[3];
    int main()
    {
        int b,d,m;
        scanf("%d %d %d",&s[0],&s[1],&s[2]);
        int a=s[0]+s[1]+s[2]+4;//一个小技巧,加上4以后能够将不足5个所需的柜子书加上!

    b=a/5; scanf("%d %d %d",&y[0],&y[1],&y[2]); int c=y[0]+y[1]+y[2]+9;//同上 d=b/10; scanf("%d",&m); if(b+d>m) printf("NO "); else printf("YES "); return 0; }



  • 相关阅读:
    执行一个外部程序并等待他的结束
    打开WORD文档时提示“word无法启动转换器mswrd632 wpc”的解决方法
    WaitForSingleObject的用法
    Delphi CreateProcess WIN32API函数CreateProcess用来创建一个新的进程和它的主线程,这个新进程运行指定的可执行文件
    webservices传base64字串
    webservices传文件
    webservices 字节数组 Base64编码
    内存映射大文件
    文件分割合并
    move
  • 原文地址:https://www.cnblogs.com/blfbuaa/p/6773620.html
Copyright © 2011-2022 走看看