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题意 : 给你一个 N 求 C(N,1 to N ) 的LCM;
有定理: (不会推)
g (n ) = C(N,1 to N ) 的LCM,
f (n ) = Lcm ( 1, to n)
g ( n ) = f(n + 1 ) / ( n + 1 );
f ( n ) = f (n –1 ) * p (( if n 为 质数 p 的k 次方数)
or = f (n – 1 );
根据上式就可以很快得出 g (n )了;
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; const int maxn = 1e6 +131; const ll mod = 1e9 + 7; int Prime[maxn]; bool Jug[maxn]; int Cnt; void Make() { Cnt = 0; for(int i = 2; i < maxn; ++i) { if(Jug[i] == false) { Prime[Cnt++] = i; for(int j = i; j < maxn; j += i) Jug[j] = true; } } } ll Fun[maxn]; void MK() { Fun[1] = 1; for(int i = 2; i < maxn; ++i) { int tmp = i; bool jj = true; for(int j = 0; j <Cnt && Prime[j] <= tmp; ++j) { if(tmp % Prime[j] == 0) { while(tmp % Prime[j] == 0) tmp /= Prime[j]; if(tmp > 1) jj = false; break; } } if(jj) Fun[i] = Fun[i-1] * i % mod; else Fun[i] = Fun[i-1]; } } ll ppow(ll a, ll b) { ll ret = 1; while(b) { if(b & 1) ret = ret * a % mod; a = a * a % mod; b >>= 1; } return ret; } int main() { Make(); //MK(); int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); ll ans = 1; for(int i = 0; i < Cnt && Prime[i] <= n; ++i) { ll tmp = Prime[i]; while(tmp <= n) { if((n+1) % tmp != 0) ans = ans * Prime[i] % mod; tmp = tmp * Prime[i]; } } printf("%lld\n",ans); } return 0; }
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原文地址:http://www.cnblogs.com/aoxuets/p/4746278.html