POJ 1091 跳蚤

简单分析题意，是要满足a1 * x1 + a2 * x2 + a3 * x3 + ...... am * xm = 1， 就是要求x1, x2, x3 , .....xm的公约数为1, 否则不可能满足。能够组成的数的组合总数为m ^ n种， 然后求有公约数不只有1的数的组合，相减即为结果。

将m分解为质因数的乘积，例如20 = 2 ^ 2 * 5；然后公约数有2的数组个数为(m / 2) ^ n， 总数利用容斥原理求解。 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

typedef long long ll;

const int LEN = 1e4 + 10;

bool is[LEN];
ll prime[LEN];
int cnt;
vector <int> v;

ll qpow(int n, int m) {
ll tmp = n;
ll ret = 1;
while (m) {
if (m & 1) {
ret *= tmp;
}
m >>= 1;
tmp = tmp * tmp;
}
return ret;
}

void getPrime() {
cnt = 0;
memset(is, 0, sizeof(is));
for (int i = 2; i < LEN; i++) {
if (is[i])
continue;
prime[cnt++] = i;
for (int j = i * i; j < LEN; j += i) {
is[j] = 1;
}
}
}

void divide(int n) {
v.clear();
for (int i = 0; i < cnt && prime[i] <= n; i++) {
if (n % prime[i] == 0) {
v.push_back(prime[i]);
}
while (n % prime[i] == 0)
n /= prime[i];
}
if (n > 1)
v.push_back(n);
}

ll getNum(int n, int m) {
ll g = 0;
int tsize = v.size();
int Maxs = (1 << tsize);

for (int s = 1; s < Maxs; s++) {
ll total = 0, pro = 1;
for (int i = 0; i < tsize; i++) {
if (s & (1 << i)) {
pro *= v[i];
total++;
}
}
if (total & 1)
g += qpow(m / pro, n);
else
g -= qpow(m / pro, n);
}
return g;
}

int main() {
getPrime();

int n, m;
while (~scanf("%d%d", &n, &m)) {
divide(m);
ll g = getNum(n, m);
ll ans = qpow(m, n) - g;

printf("%lld\n", ans);

}

return 0;
}