codeforces 842 A - Kirill And The Game

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本文介绍了一种算法，用于解决游戏中购买特定效率药剂的问题。玩家希望找到经验值与成本比值为k的药剂。文章通过遍历成本区间，判断是否存在符合条件的药剂。

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Topic:

Kirill plays a new computer game. He came to the potion store where he can buy any potion. Each potion is characterized by two integers — amount of experience and cost. The efficiency of a potion is the ratio of the amount of experience to the cost. Efficiency may be a non-integer number.

For each two integer numbers a and b such that l ≤ a ≤ r and x ≤ b ≤ y there is a potion with experience a and cost b in the store (that is, there are (r - l + 1)·(y - x + 1) potions).

Kirill wants to buy a potion which has efficiency k. Will he be able to do this?

Input:

First string contains five integer numbers l, r, x, y, k (1 ≤ l ≤ r ≤ 107, 1 ≤ x ≤ y ≤ 107, 1 ≤ k ≤ 107).

Output:

Print “YES” without quotes if a potion with efficiency exactly k can be bought in the store and “NO” without quotes otherwise.

You can output each of the letters in any register.

Example:

Input

1 10 1 10 1

Output

YES

Input

1 5 6 10 1

Output

NO

题意:

 药剂 都有 经验值 和成本   经验值和成本的比值,就是药剂的效率(效率可能是小数).
 一种药剂,在商店里 经验值为 a(l<=a<=r) ,成本为 b(x<=b<=y) 共有 (r-l+1)*(y-x+1) 个药剂.
 想买一个有效率 k 的药水。

思路:

将x至y的每一个数都乘以k只要有一个数>=l并且<=r，就可以YES，如果没有，就NO。

题解:

#include<bits/stdc++.h>
using namespace std;
int main()
{
	long long int l,r,x,y,k,i,s;
	cin>>l>>r>>x>>y>>k;
for(i=x;i<=y;i++)
{
	s=i*k;
	if(s>=l&&s<=r)
	{
		cout<<"YES"; 
		return 0;
	}
}
cout<<"NO";
	return 0;
}