codeforces-448A-Rewards

Rewards
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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:

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

第一次在codeforces上做的题
//取余问题,大水题

AC代码:

#include<stdio.h>
#include<string.h>
int main()
{
    int x,y,z,k,a,b,c,m;
    char s[5];
    scanf("%d%d%d",&x,&y,&z);
    scanf("%d%d%d",&a,&b,&c);
    scanf("%d",&k);
    ((x+y+z+4)/5+(a+b+c+9)/10)>k?m=1:m=2;
    if(m==2)
      printf("YES\n");
    else
      printf("NO\n");
    return 0;
}
时间: 2024-10-05 11:18:38

codeforces-448A-Rewards的相关文章

Codeforces Round #157 (Div. 1) C. Little Elephant and LCM (数学、dp)

C. Little Elephant and LCM time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output The Little Elephant loves the LCM (least common multiple) operation of a non-empty set of positive integers. The

Codeforces B. Taxi 算法题解

简单总结题意: 有一个数组,数组中的数值不会大于4,即只能是1,2,3,4,要把这些数值装进一个容量最大为4的容器里面,使得所用到这样的容器的个数最小. 经测试数据很大,会有10万个数据,所以这里我并不用排序数据,而是使用counting sort的思想,根据特定的数据优化,使得题解时间复杂度为O(n). 更多精彩内容:http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/ 程序如下: #include <iostream>

CodeForces 233B Non-square Equation

链接: http://codeforces.com/problemset/problem/233/B 题目: B. Non-square Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Let's consider equation: x2+s(x)·x-n=0, where x,n are positive

CodeForces:200C: Football Championship

地址链接: CF:  http://codeforces.com/problemset/problem/200/C HUST Virtual Judge: http://acm.hust.edu.cn:8080/judge/problem/viewProblem.action?id=28923 题目: C. Football Championship time limit per test 2 seconds memory limit per test 256 megabytes input s

Codeforces Round #205 (Div. 2) / 353C Find Maximum (贪心)

Valera has array a, consisting of n integers a0,a1,...,an-1, and function f(x), taking an integer from 0 to 2n-1 as its single argument. Value f(x) is calculated by formula , where value bit(i) equals one if the binary representation of number xconta

Codeforces Round #201 (Div. 1) / 346A Alice and Bob:数论&amp;amp;想法题

A. Alice and Bob http://codeforces.com/problemset/problem/346/A time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output It is so boring in the summer holiday, isn't it? So Alice and Bob have inven

算法:CodeForces #196(Div. 2) 337D Book of Evil(树形dp)

题意 给一棵n个结点的树,任意两个节点的距离是指连接两点的最短的边数 在树上的某个结点 有一个"恶魔之书",这本书会让距离它d以内的节点都受到影响 已知有m个节点收到了影响,问最多有几 个结点可能放着"恶魔之书"? 思路 要判断某个点是不是放着书,就要判断这个点的周围d距离以 内是否包含所有受影响的m节点 而如果某个节点距离最远的那个受影响节点的距离是L,如果L <= d,那 么说明所有受影响的m节点都在d以内,就可判断这个点可能放着书 那么,我们只要能够求出

Codeforces Round #100 / 140A:简单几何

http://codeforces.com/contest/140/problem/A 过大圆圆心作小圆切线即可发现规律,详见代码. 注意判相等一定要用fabs!!! 完整代码: /*30ms,0KB*/ #include<bits/stdc++.h> using namespace std; int main() { int n, R, r; double a; scanf("%d%d%d", &n, &R, &r); if (r > R |

CodeForces 275C 贪心

A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x<y) from the set, such that y=x·k. You're given a set of n distinct posi

算法题:CodeForces 236B

链接: http://www.codeforces.com/problemset/problem/236/B 题目: B. Easy Number Challenge time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Let's denote d(n) as the number of divisors of a positiv