【题目链接】
【算法】
令n!=z,因为1 / x + 1 / y = 1 / z,所以x,y>z,不妨令y = z + d
则1 / x + 1 / (z + d) = 1 / z
1 / x = 1 / z - 1 / (z + d)
1 / x = d / (z + d)z
x = z(z + d) / d = z^2 / d + z
因为x是正整数,所以z^2 / d是正整数,所以d | z^2
问题就转化为了求z^2的约数个数
约数个数定理 x = p1^k1p2^k2....pn^kn,(p1,p2,....pn)为质数,x的约数个数为(k1+1)(k2+1)...(kn+1)
分解质因数即可
【代码】
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int MAXN = 1e6; const int MOD = 1e9 + 7; ll N,i,k,ans=1,tot=0; ll prime[MAXN+10],f[MAXN+10],sum[MAXN+10]; template <typename T> inline void read(T &x) { ll f=1; x=0; char c = getchar(); for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; } for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> inline void write(T x) { if (x < 0) { putchar('-'); x = -x; } if (x > 9) write(x/10); putchar(x%10+'0'); } template <typename T> inline void writeln(T x) { write(x); puts(""); } inline void sieve(ll n) { ll i,j,tmp; for (i = 2; i <= n; i++) { if (!f[i]) prime[++tot] = f[i] = i; for (j = 1; j <= tot; j++) { tmp = i * prime[j]; if (tmp > n) break; f[tmp] = prime[j]; if (f[i] == prime[j]) break; } } } inline void calc(ll x) { while (x != 1) { sum[f[x]]++; x /= f[x]; } } int main() { read(N); sieve(N); for (i = 1; i <= N; i++) calc(i); for (i = 1; i <= tot; i++) ans = ans * (2 * sum[prime[i]] + 1) % MOD; writeln(ans); return 0; }