#include <iostream>
#include <cstdio>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>
#include <stdlib.h>
#include <stdio.h>
#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int
#define l(x) x<<1
#define r(x) x<<1|1
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;
int n, m;
vector<int> x;
void getprime(int m)
{
//得到m的质因子
x.clear();
for (int i = 2; i*i <= m; ++i)
if (m%i == 0)
{
x.push_back(i);
while (m%i == 0) m /= i;
}
if (m != 1) x.push_back(m);
}
int main() {
while (scanf("%d%d", &n, &m) == 2) {
getprime(m);//分解m
ll ans = pow(m, n);//ans为m^n
for (int i = 1; i < (1 << x.size()); ++i) {//m素因子个数为t。则i<2^t。即m素因子集合的子集个数
int k = 1, cnt = 0;
for (int j = 0; j < x.size(); ++j)
if (i&(1 << j))
k *= x[j], ++cnt;//k为取出的这个子集的值的积。cnt表示子集中元素的个数
if (cnt & 1) //等价于cnt%2==1。取出个数为奇数时减,取出个数为偶数时加
ans -= pow(m / k, n);
else
ans += pow(m / k, n);
}
printf("%lld\n", ans);
}
return 0;
}
poj 1091 (容斥原理)
最新推荐文章于 2021-05-30 09:54:28 发布