ZOJ 2674 Strange Limit

最新推荐文章于 2019-01-31 10:08:18 发布

aozil_yang

最新推荐文章于 2019-01-31 10:08:18 发布

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分类专栏：数学 ZOJ 文章标签： c语言 zoj

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本文探讨如何解决快速幂序列在阶乘取模情况下的极限问题，通过多套快速幂操作，实现对无限序列的逼近，进而求得极限值。此方法巧妙地结合了数学递归和模运算的特性，简化了复杂序列的求解过程。

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Strange Limit

Time Limit: 5 Seconds Memory Limit: 32768 KB

Consider sequence a_n defined with the following recurrence:

a₁= p,
a_n+1= p^{^a^_n} for n >= 1,

where p is some prime number. Let

b_n = a_n mod m!,

where m! denotes m factorial, that is m! = 1 · 2 · ... · m.

It may seem strange, but for all p and all m the sequence b_n has limit as n . Your task is to find it. Given p and m, find

Input

There are several test cases in the input. Each case contains p and m (2 <= p, m <= 12, p is prime). There is am empty line between each case.

Output

Output the limit requested. There should be am empty line between each case.

Example

Input	Output
2 2	0
2 3	4
3 8	30267

Source: Andrew Stankevich's Contest #8

大体题意：

给你两个数p，m，计算bn的极限，其中bn等于an对m的阶乘取模，而an = p^p^p...

其实是水过的，n趋向无穷大，就是有无数个p，那么只要让n稍微大一点，就差不多不变了，就是接近无穷大了，所以直接快速幂对m阶乘取模不就可以啦！就是多套几个快速幂就可以求出“极限”，打表就行了。！

正解学会了在修改！

代码如下：

#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
ll fac(ll n){if (!n)return 1;return n*fac(n-1);}// 求阶乘
ll my_pow(ll a,ll n,ll mod){ll ans = 1;while(n){if (n % 2)ans = (ans * a)%mod;a = (a%mod*a%mod)%mod;n/=2;}return ans;}// 快速幂！对m的阶乘取模
int main()
{
    ll m1=3,m2=4,n,k, p[6] = {2,3,5,7,11},a[50][50];
    for (int k = 0; k < 5; ++k){
        for (ll m2 = 2; m2 <=12; ++m2){
            m1=p[k];
            a[m1][m2]=my_pow(m1,my_pow(m1,my_pow(m1,my_pow(m1,my_pow(m1,my_pow(m1,my_pow(m1,m1,fac(m2)),fac(m2)),fac(m2)),fac(m2)),fac(m2)),fac(m2)),fac(m2));//多套几个快速幂 就是“无穷大”了！
        }
    }
    int cnt = 0;
    while(cin >> n >> k){
        cnt++;
        if (cnt > 1)cout << endl;//注意格式！
        cout << a[n][k] << endl;
    }
    return 0;
}