题意:求1-n!里与m!互质的数有多少？(m<=n<=1e6).
因为n!%m!=0,所以题目实际上求的是phi(m!)*n!/m!.
预处理出这些素数的逆元和阶乘的模即可。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <bitset> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-8 # define MOD 30031 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } const int N=10000005; //Code begin... int fac[N], pri[N], inv[N], ans[N]; bool vis[N]; int P, tot; void init(int n) { fac[1]=1; FOR(i,2,n) fac[i]=(LL)fac[i-1]*i%P; inv[1]=1; FOR(i,2,n){ if(!vis[i])pri[++tot]=i; for(int j=1; pri[j]*i<=n&&j<=tot; ++j){ vis[pri[j]*i]=1; if(i%pri[j]==0) break; } } for(int i=2; i<=n&&i<P; ++i) inv[i]=(P-(LL)P/i*inv[P%i]%P); ans[1]=1; FOR(i,2,n) { ans[i]=ans[i-1]; if(!vis[i])ans[i]=(LL)ans[i]*(i-1)%P*inv[i%P]%P; } } int main() { int T, n, m; scanf("%d%d",&T,&P); init(N-5); while(T--){ scanf("%d%d",&n,&m); printf("%d ",(LL)fac[n]*ans[m]%P); } return 0; }