题目大意

求
\[\sum_{i = 1}^{N!} [gcd(i, M!) = 1]\]

题解

显然，题目就是求
\[N!(1-\frac{1}{p_1})(1-\frac{1}{p_2})...\]
即
\[N!\prod(p_i - 1)(\prod p_i)^{-1}\]
预处理一下，都是线性复杂度。
注意：

1. N=1的情况
2. long long

所以，数论题一定要注意各种特殊情况和longlong

代码

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 10000110;
const int N = 10000010;
int n, m, t, r;
int prime[maxn], check[maxn], d[maxn];
int prd1[maxn], prd2[maxn], fact[maxn];
int tot = 0;

inline int read() {
char c = getchar();
int f = 1, x = 0;
while (!isdigit(c)) {
if (c == '-')
f = -1;
c = getchar();
}
while (isdigit(c))
x = x * 10 + c - '0', c = getchar();
return x * f;
}

inline void get_prime(int n) {
memset(check, 0, sizeof(check));
for (int i = 2; i <= n; i++) {
if (!check[i])
prime[tot++] = i;
for (int j = 0; j < tot; j++) {
if (i * prime[j] > n)
break;
check[i * prime[j]] = 1;
if (i % prime[j] == 0)
break;
}
}
}
inline void get_prd(int p) {
// get prd (p_i - 1) and prd (p_i);
prd1[0] = 1;
prd2[0] = 2;
for (int i = 1; i <= tot; i++) {
prd1[i] = (((ll)prime[i] - 1) % p * (ll)prd1[i - 1]) % p;
prd2[i] = (ll)(prime[i] % p * (ll)prd2[i - 1]) % p;
}
}
inline void init() {

for (int i = 0; i < tot; i++) {
for (int j = prime[i]; j < prime[i + 1]; j++)
d[j] = i;
}
fact[0] = fact[1] = 1;
for (int i = 2; i <= N; i++)
fact[i] = (ll)(fact[i - 1] * (ll)i) % r;
}
int pow(int a, int b, int p) {
int x = 1;
int c = b;
while (c) {
if (c & 1)
x = (ll)((ll)x * a) % p;
a = (ll)((ll)a * a) % p;
c >>= 1;
}
return x;
}
int inv(int a, int p) { return pow(a, p - 2, p); }
int main() {
scanf("%lld %lld", &t, &r);
get_prime(N);
get_prd(r);
init();
while (t--) {
n = read();
m = read();
if (m == 1) {
printf("%lld\n", fact
);
continue;
}
int ans = ((ll)((ll)fact
 * prd1[d[m]]) % r * (ll)inv(prd2[d[m]], r)) % r;
printf("%d\n", ans);
}
}