A.Accumulator--简单数学

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分类专栏： # 2019湖南多校第五场 # 基本数学定理运用

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博客围绕手机电量消耗问题展开，Anna手机电量到20%时消耗速度变慢两倍，已知她从家到见面处用时t及到达时电量p，需计算到达后多久手机没电。这是方程问题，分电量大于等于20%和小于20%两种情况讨论，并给出了相关方程及AC代码。

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Accumulator

Time Limit: 3 Sec Memory Limit: 512 Mb

题目链接http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2308

Description

Anna loves her cell phone and becomes very sad when its battery level drops to 0 percent.

In normal mode, Anna’s phone battery drains at a constant speed. When the battery level reaches 20 percent, the phone automatically switches to eco mode. In eco mode, the battery drains two times slower than in normal mode.

Alex has invited Anna for a date. Anna needs t minutes to get from her home to the meeting place. When Anna leaves home, her phone’s battery level is 100 percent. At the moment she reaches the meeting place, the battery level will be p percent.

Alex wonders for how long Anna will be in a good mood after they meet. Help him solve this problem!

Input

The only line of the input contains two integers t and p — time Anna needs to get from her home to the meeting place, in minutes, and the battery level of her phone at the moment of meeting, in percent (1 ≤ t ≤ 360; 1 ≤ p ≤ 99).

Output

Output a single real number — time since the moment of meeting before Anna’s phone runs out of battery, in minutes.

Your answer will be considered correct if its absolute or relative error doesn’t exceed 10^-4

Sample Input

30 70

Sample Output

90.0

emmm，题目大意：A要到B的家里，A的手机电量到百分之二十的时候消耗的速度比正常的慢两倍，当A离开家时，她手机的电池电量是100％。在她到B的那一刻，电池电量将是p％，用时t。请问到B家中多久后A的手机电量为0.

其实这就是个方程问题，稍微再草稿纸上演算一下也就出来了，分情况讨论，当电量大于等于20%的时候很简单就不多说了。当小于20%的时候有方程：
v1* t1+v2* t2=100-p—> 2v2* t1+v2* t2=100-p-----> v2(2t1+t2)=100-p
即：v2* t+ v2* t1=100-p，又v2=1/2* v1所以v2* t=100-p-80/2

以下是AC代码：

#include <bits/stdc++.h>
using namespace std;
int main()
{
	int t,p;
	scanf ("%d%d",&t,&p);
	if (p>=20){
		double rate=1.0*(100-p)/t;
		double ans=1.0*(p-20)/rate;
		ans+=1.0*20/(rate/2);
		printf ("%.6f\n",ans);
	}
	else {
		double rate=1.0*(60-p)/t;
		double ans=1.0*p/rate;
		printf ("%.6f\n",ans);
	}
	return 0;
}