Rewards - #256 (Div. 2) A (448A)水题

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本文探讨了如何合理分配奖杯和奖牌到多个货架上，确保每个货架不同时存放奖杯和奖牌，且限制奖杯不超过5个，奖牌不超过10个。通过实例分析，提供了解决方案。

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 a1, a2 and a3 (0 ≤ a1, a2, a3 ≤ 100). The second line contains integers b1, b2 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：上面三个是奖杯，下面三个是奖牌，一个橱子最多放5个奖杯或10个奖牌，问现有的橱子是否够用……………一开始理解成是否能把在把所有物品都放进橱子的情况下，使橱子填满……。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
int main()
{ int n,m,i,j,k,num1=0,num2=0,mi=0;
  for(i=1;i<=3;i++)
  { scanf("%d",&k);
    num1+=k;
  }
  for(i=1;i<=3;i++)
  { scanf("%d",&k);
    num2+=k;
  }
  mi=num1/5+num2/10;
  num1%=5;
  num2%=10;
  if(num1)
   mi++;
  if(num2)
   mi++;
  scanf("%d",&n);
  if(mi<=n)
   printf("YES\n");
  else
   printf("NO\n");
}