POJ 1091 跳蚤

　　HNOI的题。。。蛋疼的没有想到，看了discuss。。。

　　已知N+1个数，x1, x2, .. xn, m；使得gcd(x1, x2, ..., xn) = 1(mod m)。也就是说gcd(x1, x2,..., xn, m) = 1

需要找x1, x2, ... ,xn这样的序列多少个。其实就是找到与m互素的数构造这个集合就可以。。。

首先对m进行分解质因子，可以用容斥原理找到与m不互素的数构成的集合，用m^n减掉就可以了；

 ps:貌似这题不用高精度。

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))
#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))
#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))
#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define Read()  freopen("data.in", "r", stdin)
#define Write() freopen("data.out", "w", stdout);

typedef long long LL;
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;


using namespace std;

const int N = 20010;

int prime[N], cnt;
bool vis[N];

int p[N], num;

void get_prime() {
    CL(vis, true);
    int i, j;
    for(i = 2; i < N; ++i) {
        for(j = i*i; j < N; j += i) {
            vis[j] = false;
        }
    }
    cnt = 0;
    for(i = 2; i < N; ++i) {
        if(vis[i])  prime[cnt++] = i;
    }
}

void get_p(int m) {
    int i;
    num = 0;
    for(i = 0; i < cnt && prime[i] <= m; ++i) {
        if(m%prime[i] == 0) {
            p[num++] = prime[i];
            while(m%prime[i] == 0) { m /= prime[i]; }
        }
    }
    if(m != 1)  p[num++] = m;
}

LL exp(LL a, LL b) {
    LL res = 1;
    while(b--)  res *= a;
    return res;
}

int main() {
    //Read();

    int n, m, j, bit;
    get_prime();
    while(~scanf("%d%d", &n, &m)) {
        get_p(m);
        LL res = 0, sum, i;

        for(i = 1; i < (1<<num); ++i) {
            bit = 0; sum = 1;
            for(j = 0; j< num; ++j) {
                if(i&(1<<j)) {
                    bit++; sum *= p[j];
                }
            }
            if(bit&1)   res -= exp(m/sum, n);
            else    res += exp(m/sum, n);
        }
        LL ans = exp(m, n);
        //printf("%lld\n", ans);
        printf("%lld\n", ans + res);
    }
    return 0;
}