A. Kirill And The Game_time limit per test 1.5 seconds memory limit per t-优快云博客

本文介绍了一个计算机游戏中的挑战任务，玩家需要在特定的数值范围内找到符合条件的药水购买方案，以实现给定的经验与成本比例。

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.

Examples
input
1 10 1 10 1
output
YES
input
1 5 6 10 1
output
NO

题意：

给你两个区间[L,R].[X,Y]，和一个整数k。求是否能从第一个区间中取一个整数a，第二个区间取一个整数b，使得a/b恰好等于k。

思路：

由于a,b,k为整数，为避免a/b自动取整问题，所以要判断b*k等于a。但是不能直接判断[k*x,k*y]与[[l,r]是否有交集来判断存不存在，，因为这两个区间是不连续的，只能取整数。

#include <iostream>
#include <bits/stdc++.h>

using namespace std;
//c
int main()
{
    long long i,r,x,y,k;
    int fg=0;
    cin>>i>>r>>x>>y>>k;
    for(int j=x; j<=y; j++)
    {
        if(j*k>=i&&j*k<=r)
            fg=1;
    }
    if(fg)
        cout<<"YES\n";
    else
        cout<<"NO\n";
    return 0;
}