Codeforces Round #430 A. Kirill And The Game

最新推荐文章于 2018-03-05 11:49:00 发布

爱不到要偷

最新推荐文章于 2018-03-05 11:49:00 发布

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原文链接：http://www.cnblogs.com/Yuzao/p/7451716.html

本文介绍了一道关于游戏内购效率的问题，玩家需要在给定的经验值和成本范围内找到符合条件的药水购买方案。通过枚举成本并判断经验是否符合预期比例的方法解决了这一问题。

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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.

题目大意：两个数字a,b，给出a，b的范围，询问是否存在a/b=k
解题报告：直接枚举b，判断k×b是否在a的范围内即可

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
#define il inline
#define iter iterator
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
void work()
{
    int l,r,x,y,k;
    scanf("%d%d%d%d%d",&l,&r,&x,&y,&k);
    long long tmp=0;
    for(int i=x;i<=y;i++){
        tmp=(long long)k*i;
        if(tmp>=l && tmp<=r){
            puts("YES");
            return ;
        }
    }
    puts("NO");
}

int main()
{
    work();
    return 0;
}

转载于:https://www.cnblogs.com/Yuzao/p/7451716.html