2017年第0届浙江工业大学之江学院程序设计竞赛决赛—D qwb与神奇的序列

Description

Input

Output

Sample Input

1 2 2

Sample Output

15

思路：规律题，首先容易推出ans[i] = ans[i-1]*3-(x+y), 再根据等比数列公式可以得到规律第n次操作的和

为：(3^n+1)/2 * (x+y);

快速幂做法需要几个注意的点：

1. 不能在求3^n的时候直接取模，之后/2%mod

2. 看到除法就想逆元是不可行的，因为2和mod不互质，所以逆元不能用。

3. 可以根据公式：(a/b)%mod = a%(b*mod)/b%mod, 所以就可以在快速幂求3^n的时候%2e8，求完之后

再/2%mod*(x+y)%mod，即可得到正确答案。

根据ans[i] = ans[i-1]*3-(x+y)还可以得到矩阵快速幂做法，

细节稍微推推就出来了。

Code1:

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int mod = 1e8;
LL a, b, n;
LL quickM(LL a, LL b)
{
	LL ans = 1, base = a;
	while(b)
	{
		if(b&1) ans = ans*base%(mod*2);
		base = base*base%(mod*2);
		b >>= 1;
	}
	return ans;
}
int main()
{
	while(~scanf("%lld %lld %lld", &a, &b, &n))
	{
		x = (a+b)%mod;
		ans = (quickM(3, n)+1)/2%mod*x%mod;
		printf("%lld\n", ans);
	}
	return 0;
}

Code2:

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const LL mod = 1e8;
struct node
{
	LL e11, e12, e13, e21, e22, e23, e31, e32, e33;
};
LL x, y, n;
node multi(node a, node b)
{
	node ans;
	ans.e11 = (a.e11*b.e11 + a.e12*b.e21 + a.e13*b.e31)%mod;
	ans.e12 = (a.e11*b.e12 + a.e12*b.e22 + a.e13*b.e32)%mod;
	ans.e13 = (a.e11*b.e13 + a.e12*b.e23 + a.e13*b.e33)%mod;
	ans.e21 = (a.e21*b.e11 + a.e22*b.e21 + a.e23*b.e31)%mod;
	ans.e22 = (a.e21*b.e12 + a.e22*b.e22 + a.e23*b.e32)%mod;
	ans.e23 = (a.e21*b.e13 + a.e22*b.e23 + a.e23*b.e33)%mod;
	ans.e31 = (a.e31*b.e11 + a.e32*b.e21 + a.e33*b.e31)%mod;
	ans.e32 = (a.e31*b.e12 + a.e32*b.e22 + a.e33*b.e32)%mod;
	ans.e33 = (a.e31*b.e13 + a.e32*b.e23 + a.e33*b.e33)%mod;
	return ans;
}
LL quickM(LL b)
{
	node base, ans;
	base.e11 = 3, base.e12 = 0, base.e13 = -1;
	base.e21 = 0, base.e22 = 3, base.e23 = -1;
	base.e31 = 0, base.e32 = 0, base.e33 = 1;
	ans.e11 = 1, ans.e12 = 0, ans.e13 = 0;
	ans.e21 = 0, ans.e22 = 1, ans.e23 = 0;
	ans.e31 = 0, ans.e32 = 0, ans.e33 = 1;
	while(b)
	{
		if(b&1) ans = multi(ans, base);
		base = multi(base, base);
		b >>= 1;
	}
	return (ans.e21*2+ans.e22*1+ans.e23+mod)%mod;
}
int main()
{
	while(~scanf("%lld %lld %lld", &x, &y, &n))
	{
		printf("%lld\n", (x+y)%mod*quickM(n)%mod);
	}
	return 0;
}

------------------------------优雅的分割线------------------------------

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const LL mod = 1e8;
struct node
{
	LL m[4][4];
};
LL x, y, n;
node multi(node a, node b)
{
	node ans;
	memset(ans.m, 0, sizeof ans.m);
	for(int i = 1; i <= 3; ++i)
		for(int j = 1; j <= 3; ++j)
			for(int k = 1; k <= 3; ++k)
				ans.m[i][j] = (ans.m[i][j]+a.m[i][k]*b.m[k][j])%mod;
	return ans;
}
LL quickM(LL b)
{
	node base, ans;
	base.m[1][1] = 3, base.m[1][2] = 0, base.m[1][3] = -1;
	base.m[2][1] = 0, base.m[2][2] = 3, base.m[2][3] = -1;
	base.m[3][1] = 0, base.m[3][2] = 0, base.m[3][3] = 1;
	ans.m[1][1] = 1, ans.m[1][2] = 0, ans.m[1][3] = 0;
	ans.m[2][1] = 0, ans.m[2][2] = 1, ans.m[2][3] = 0;
	ans.m[3][1] = 0, ans.m[3][2] = 0, ans.m[3][3] = 1;
	while(b)
	{
		if(b&1) ans = multi(ans, base);
		base = multi(base, base);
		b >>= 1;
	}
	return (ans.m[2][1]*2+ans.m[2][2]*1+ans.m[2][3]+mod)%mod;
}
int main()
{
	while(~scanf("%lld %lld %lld", &x, &y, &n))
	{
		printf("%lld\n", (x+y)%mod*quickM(n)%mod);
	}
	return 0;
}

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