uva 10692 高次幂取模

Huge Mod

Input: standard input

Output: standard output

Time Limit: 1 second

The operator for exponentiation is different from the addition, subtraction, multiplication or division operators in the sense that the default associativity for exponentiation goes right to left instead of left to right. So unless we mess it up by placing parenthesis,  should mean  not . This leads to the obvious fact that if we take the levels of exponents higher (i.e., 2^3^4^5^3), the numbers can become quite big. But let's not make life miserable. We being the good guys would force the ultimate value to be no more than 10000.

Given a₁, a₂, a₃, ... , a_N and m(=10000) you only need to compute a₁^a₂^a₃^...^a_N mod m.

 

Input

There can be multiple (not more than 100) test cases. Each test case will be presented in a single line. The first line of each test case would contain the value for M(2<=M<=10000). The next number of that line would be N(1<=N<=10). Then N numbers - the values for a₁, a₂, a₃, ... , a_Nwould follow. You can safely assume that 1<=a_i<=1000. The end of input is marked by a line containing a single hash ('#') mark.

 

 

Output

For each of the test cases, print the test case number followed by the value of a₁^a₂^a₃^...^a_Nmod m on one line. The sample output shows the exact format for printing the test case number.

 

 

Sample Input	Sample Output
10 4 2 3 4 5 100 2 5 2 53 3 2 3 2 #	Case #1: 2 Case #2: 25 Case #3: 35

 题目大意：求一个数((((a^b)^c)^d)^e)..... Mod m的值

幂太huge了，上界是1000^1000^1000^1000^1000^1000^1000^1000^1000,暴力快速幂模肯定行不通，因为幂是多少都难的计算。有公式a^x=a^(x%phi(c)+phi(c)) (mod c),所以可以用递归方法求解。

AC代码：

#include <iostream>
#include <cstdio>
#include <string>
using namespace std;
int phi[10010];
int f[20],n;
string m;

void init()
{
    int i;
    for(i=2;i<=10000;i++) phi[i]=0;
    phi[1]=1;
    for(i=2;i<=10000;i++) 
        if(!phi[i])
       for(int j=i;j<=10000;j+=i)
       {
           if(!phi[j]) phi[j]=j;
           phi[j]=phi[j]/i*(i-1);
       }
}

int montgomery(int a,int b,int c)
{
    int t=1;
    while(b)
    {
        if(b%2)
            t=t*a%c;
        b/=2;
        a=a*a%c;
    }
    return t;
}

int dfs(int now,int mod)
{
    if(now==n-1)
    {
        return f[now]%mod;
    }
    int t=dfs(now+1,phi[mod]);
    int ans=montgomery(f[now],t+phi[mod],mod);
    return ans;
}
int main()
{
    init();
    int    i,ret,kase=1;
    while(cin>>m,m!="#")
    {
        ret=0;
        for(i=0;i<m.size();i++)
        ret=ret*10+m[i]-'0';
        cin>>n;
        for(int i=0;i<n;i++)
            scanf("%d",f+i);
        cout<<"Case #"<<kase++<<": ";
        printf("%d
",dfs(0,ret));
    }
    return 0;
}