kari492 欧拉函数

492. Goblin's Appreciation

时间限制 1000 ms  内存限制 65536 KB

    They give you a lot of pairs of numbers, denoted by K,N. for every input, you should tell them the sum of k^gcd(i,n), for i = 1~n.
    For the answer is so large, you can give them the answer mod 23333.

输入格式

Given T, the number of cases. Then, for next T lines, there are two numbers K,N.(1 <= k <= 10^4, 1 <= n <= 10^8)

 输出格式

For every case, print one line, containing the answer mod 23333.

 输入样例

3
3 4
2 6
100 50000000

输出样例

96
84
3014

分析：10的8次方，直接算肯定超时。我们考虑与n的gcd相同的数的个数，gcd(i,n)=x————gcd(i/x,n/x)=1————即小于等于n/x且与x互质的数的个数，也就是欧拉函数。phi=n(1-1/p1)(1-1/p2)……(1-1/pn).

计算phi[n]，直观感觉打表比每个暴力算要快，但是实际上我们只需要计算n的约数的phi值，个数远小于n本身（尤其是n很大时）。所以枚举n的约数，再逐个计算更快。而且打表为O(nlogn),10的8次方超时。

代码：
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#define MOD 23333
using namespace std;
long long  phi[1000001];
long long  power[60];
void pp(long long k)
{
    for(int i=1;i<=32;i++)
    {
        power[i]=k;
        k=k*k%MOD;
    }
}
long long  mod_power(long long k,long long n)
{
    long long res=1;
    int i=1;
    while(n)
    {
        if(n&1)
            res=res*power[i]%MOD;
        n>>=1;
        i++;
    }
    return res;
}
//O(根n)的复杂度
int euler_phi(int n)
{
    int m=(int)sqrt(n+0.5);
    int ans=n;
    for(int i=2;i<=m;i++)
    {
        if(n%i==0)
        {
            ans=ans/i*(i-1);
            while(n%i==0)   //将该约数除尽
                n/=i;
        }
    }
    if(n>1)
        ans=ans/n*(n-1);
    return ans;
}
int main()
{
    long long t,n,i,j,k,sum;
    scanf("%lld",&t);
    while(t--)
    {
        sum=0;
        scanf("%lld %lld",&k,&n);
        pp(k);
        int m=(int)sqrt(n+0.5);
        for(i=1;i<=m;i++)
        {
            if(n%i==0)
            {
                sum=(sum+mod_power(k,n/i)*euler_phi(i)%MOD)%MOD;
                if(n/i!=i)
                    sum=(sum+mod_power(k,i)*euler_phi(n/i)%MOD)%MOD;
            }
        }
        printf("%lld\n",sum);
    }
    return 0;
}