#include <iostream>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <algorithm>
#include <cmath>
using namespace std;
int main()
{
double a[10],b[10],suma=0,sumb=0;
int n,i,j;
for(i=0; i<3; i++)
{
cin>>a[i];
suma+=a[i];
}
for(i=0; i<3; i++)
{
cin>>b[i];
sumb+=b[i];
}
cin>>n;
suma=ceil(suma/5);
sumb=ceil(sumb/10);
if(suma+sumb>n)
cout<<"NO"<<endl;
else cout<<"YES"<<endl;
return 0;
}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.
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.
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).
1 1 1 1 1 1 4
YES
1 1 3 2 3 4 2
YES
1 0 0 1 0 0 1
NO
探讨如何在有限的柜架中合理摆放不同数量的奖杯和奖牌,确保每层只放置一种奖励,同时遵守奖杯不超过5个、奖牌不超过10个的规则。
524

被折叠的 条评论
为什么被折叠?



