描述 Description

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 a3 third 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:

1.any shelf cannot contain both cups and medals at the same time;

2.no shelf can contain more than five cups;

3.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.

输入格式 InputFormat

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.

输出格式 OutputFormat

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).

样例输入 SampleInput

1 1 1
1 1 1

样例输出 SampleOutput


Codeforces 448A

代码 Code

水。太弱了只能继续在 div2 里混。。。

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int inf=0x7fffffff;
int i,j,t,n,m,l,r,k,z,y,x;
int a[4],b[4];
int suma,sumb;
int main()
    for (i=1;i<=3;i++) scanf("%d",&a[i]),suma+=a[i];
    for (i=1;i<=3;i++) scanf("%d",&b[i]),sumb+=b[i];
    if (suma%5!=0) m++;
    if (sumb%10!=0) m++;
    if (m<=n) printf("YES\n");
    else printf("NO\n");
    return 0;