G - Plant

最新推荐文章于 2024-03-10 11:18:35 发布

gtuif

最新推荐文章于 2024-03-10 11:18:35 发布

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分类专栏： Codeforces 数学快速幂

本文链接：https://blog.youkuaiyun.com/gtuif/article/details/77494271

数学同时被 3 个专栏收录

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本文介绍了一个有趣的数学问题：一种特殊三角形植物每年分裂生长的规律，并提供了一种使用快速幂运算来高效计算任意年份朝上三角形植物数量的方法。

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G - Plant

Crawling in process...

Submit Status Practice CodeForces 185A

Description

Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.

$\text{[math]}$

Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.

Input

The first line contains a single integer n(0 ≤ n ≤ 10¹⁸) — the number of full years when the plant grew.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64dspecifier.

Output

Print a single integer — the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007(10⁹ + 7).

Sample Input

Input

Output

Input

Output

Hint

The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.

题意：求出朝上的三角的个数。。。。

思路：1，1+2,1+2+3+4,1+2+3+4+5+6+7+8.。。。。。以此类推，第n个就为(2^n+1)*(2^(n-1))%mod......快速幂一发即可。。。

下面附上代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define mod 1000000007
using namespace std;
typedef long long ll;
ll Quickmod(ll a,ll b,ll c)
{
	ll ans=1;
	a=a%c;
	while(b)
	{
		if(b&1)
			ans=(ans*a)%c;
		b>>=1;
		a=(a*a)%c;
	}
	return ans;
}
int main()
{
	ll n;
	while(cin>>n)
	{
		if(n==0) puts("1");
		else
		{
			ll res=((Quickmod(2,n,mod)+1)%mod)*(Quickmod(2,n-1,mod)%mod);
			printf("%lld\n",res%mod);
		}
	}
}