CF - 630B. Moore's Law - 快速幂

最新推荐文章于 2022-05-07 15:49:00 发布

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最新推荐文章于 2022-05-07 15:49:00 发布

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分类专栏：快速幂文章标签： acm 算法 codeforces

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本文介绍了一种基于快速幂算法的摩尔定律实时计算器实现方案，该计算器能够精确估算随着时间推移集成电路中晶体管数量的增长情况。

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1.题目描述：

The city administration of IT City decided to fix up a symbol of scientific and technical progress in the city's main square, namely an indicator board that shows the effect of Moore's law in real time.

Moore's law is the observation that the number of transistors in a dense integrated circuit doubles approximately every 24 months. The implication of Moore's law is that computer performance as function of time increases exponentially as well.

You are to prepare information that will change every second to display on the indicator board. Let's assume that every second the number of transistors increases exactly 1.000000011 times.

Input

The only line of the input contains a pair of integers n (1000 ≤ n ≤ 10 000) and t (0 ≤ t ≤ 2 000 000 000) — the number of transistors in the initial time and the number of seconds passed since the initial time.

Output

Output one number — the estimate of the number of transistors in a dence integrated circuit in t seconds since the initial time. The relative error of your answer should not be greater than 10 - 6.

Examples

input

1000 1000000

output

1011.060722383550382782399454922040

2.题意概述：

求n*1.000000011 的t次方

3.解题思路：

直接快速幂搞之

4.AC代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <ctime>
using namespace std;
typedef long long ll;
#define INF 0x7fffffff
#define maxn 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e 2.718281828459
#define mod (int)1e9 + 7;
#define jin 1.000000011 
double mod_quick(double a, ll b)
{
	double ans = 1;
	while (b)
	{
		if (b & 1)
			ans *= a;
		a *= a;
		b >>= 1;
	}
	return ans;
}
int main()
{
	int n;
	ll t;
	while (scanf("%d%lld", &n, &t) != EOF)
	{
		double ans = mod_quick(jin, t);
		printf("%lf\n", n * ans);
	}
	return 0;
}