poj2154

题目意思：给定N种颜色的珠子，每种颜色个数不限，现做成长度为N的相连，问有多少种方案(只考虑旋转)。答案要mod P

思路:因为只有旋转，所以基本思路就跟poj1286差不多，而且还要简单些，

       难的话就是因为N太大了，直接算肯定TLE的。。怎么办？

      我们设 a= gcd（N， i)  ,  L= N /a,  T = i / a;  那么  gcd（L, T）= 1

      所以对于某一个T，只要T与L互质，那么 T*a与 N最大公约数必然为a

      那么就转化成找与T互质的数的个数了，这就得用到欧拉函数了。。

     最后结果为 sigma（Euler(L) * N ^ (l-1) ）% p

code:

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <cstdio>

using namespace std;
typedef __int64 LL;
int T , a[50001], bo[50001], tot;
LL n, p, sum;

void init(){
    tot = 0;
    for (int i =2; i <= 50000; ++i)
        if (!bo[i]) {
        for (int j = i+i; j <= 50000; j += i)
              bo[j] = 1;
        a[++tot] = i;
    }

}

int phi(int now){ //求欧拉函数 
    int rea = now;
    for (int i = 1; a[i]*a[i] <= now; ++i)
        if (now % a[i] == 0){
             rea -= rea / a[i];
             while (now % a[i] == 0) now /= a[i];
        }
    if (now > 1) rea -= rea / now;
    return rea;
}

LL quick(int x, int k){ //快速幂优化 
    if (k == 1) return x;
    if (k == 0) return 1;
    LL tmp = quick(x, k / 2);
    tmp = (tmp * tmp) % p;
    if (k&1) tmp = (tmp * x) % p;
    return tmp;
}

void work(int now){
     LL l = phi(now);
     sum = (sum + l * quick(n % p, n / now - 1)) % p;
}

void solve(){
     sum = 0;
     scanf("%I64d%I64d", &n, &p);
     for (int i = 1; i * i <= n; ++i)
        if (n % i == 0){
             work(i);
        //     printf("i = %d\n", i);
          //   printf("i = %d\n", n / i);
             if (i*i != n) work(n / i);
        }

     printf("%I64d\n", sum);
}

int main(){
     freopen("poj2154.in", "r", stdin);
     freopen("poj2154.out","w", stdout);
     scanf("%d", &T);
     init();
     for (int i =1; i <= T; ++i)
          solve();
     fclose(stdin); fclose(stdout);
}