Codeforces Round #256 (Div. 2/A)/Codeforces448A_Rewards(水题)

解题报告

意思就是说有n行柜子，放奖杯和奖牌。要求每行柜子要么全是奖杯要么全是奖牌，并且奖杯每行最多5个，奖牌最多10个。

直接把奖杯奖牌各自累加，分别出5和10,向上取整和N比較

#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;
}

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 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