upc 3026 Exponial

最新推荐文章于 2022-01-22 21:23:13 发布

丑心疼

最新推荐文章于 2022-01-22 21:23:13 发布

阅读量108

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Kissheart/p/9748676.html

本文探讨了如何通过欧拉降幂算法高效计算Exponial函数的值，并提供了一个具体的C++代码实现，该算法能有效解决大规模数值运算的问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Exponial

时间限制: 1 Sec 内存限制: 64 MB
提交: 229 解决: 54
[提交] [状态] [讨论版] [命题人:外部导入]

题目描述

Illustration of exponial(3) (not to scale), Picture by C.M. de Talleyrand-Périgord via Wikimedia Commons Everybody loves big numbers (if you do not, you might want to stop reading at this point). There are many ways of constructing really big numbers known to humankind, for instance:

In this problem we look at their lesser-known love-child the exponial , which is an operation deﬁned for all positive integers n as

For example, exponial(1) = 1 and

which is already pretty big. Note that exponentiation is right-associative:

.
Since the exponials are really big, they can be a bit unwieldy to work with. Therefore we would like you to write a program which computes exponial(n) mod m (the remainder of exponial(n) when dividing by m).

输入

The input consists of two integers n (1 ≤ n ≤ 10 ⁹ ) and m (1 ≤ m ≤ 10 ⁹ ).

输出

Output a single integer, the value of exponial(n) mod m.

样例输入

2 42

样例输出

题意

给一个N,M，求模M的结果。

分析

欧拉降幂的经典例题

欧拉降幂公式：

$\left\{\begin{matrix} A^{B}mod (C)=A^{Bmod(phi(C))}modC& (B<phi(C)) \\ A^{B}mod (C)=A^{Bmod(phi(C))+phi(C)}modC& (B>=phi(C)) \end{matrix}\right.$

写递归求答案就可以了。

///  author:Kissheart  ///
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<vector>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<deque>
#include<ctype.h>
#include<map>
#include<set>
#include<stack>
#include<string>
#define INF 0x3f3f3f3f
#define FAST_IO ios::sync_with_stdio(false)
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MAX=1e6+10;
long long int mod;
typedef long long ll;
using namespace std;
#define gcd(a,b) __gcd(a,b)
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
inline ll qpow(ll a,ll b,ll mod){ll r=1,t=a; while(b){if(b&1)r=(r*t)%mod;b>>=1;t=(t*t)%mod;}return r;}
//inline ll inv1(ll b){return qpow(b,mod-2);}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll r=exgcd(b,a%b,y,x);y-=(a/b)*x;return r;}
inline ll read(){ll x=0,f=1;char c=getchar();for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;for(;isdigit(c);c=getchar()) x=x*10+c-'0';return x*f;}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );

ll a[]={0,1,2,9,(1<<18)},n;
ll phi(ll n)
{
    ll ans=n;
    for (ll i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            ans-=ans/i;
            while(n%i==0)
                n/=i;
        }
    }
    if(n>1)
        ans-=ans/n;
    return ans;
}


ll f(ll n,ll m)
{
    if(m==1) return 1;
    if(n<=4)
    {
        if(a[n]>=m) return a[n]%m+m;
        return a[n];
    }
    ll exp=f(n-1,phi(m));
    return qpow(n,exp,m)+m;
}
int main()
{

    scanf("%lld%lld",&n,&mod);
    ll exp=f(n-1,phi(mod));
    ll ans=qpow(n,exp,mod)%mod;
    printf("%lld\n",ans%mod);

    return 0;
}