先令n! = a：
1 / x + 1 / y = 1 / a => x = y * a / (y - a)
再令 k = y - a：
于是x = a + a ^ 2 / k => k | a ^ 2
故等价于求a ^２的约数个数
素数筛一下什么的就好了嘛
1 /************************************************************** 2 Problem: 2721 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:588 ms 7 Memory:2408 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 12 using namespace std; 13 typedef long long ll; 14 const int N = 1000005; 15 const int Cnt = 80005; 16 const int mod = 1000000007; 17 18 int n; 19 bool F[N]; 20 int cnt, p[Cnt], c[Cnt]; 21 ll ans = 1; 22 23 int main() { 24 int i, j, t; 25 scanf("%d ", &n); 26 for (i = 2; i < N; ++i) { 27 if (!F[i]) p[++cnt] = i; 28 for (j = 1; j <= cnt && p[j] * i < N; ++j) { 29 F[p[j] * i] = 1; 30 if (p[j] % i == 0) break; 31 } 32 } 33 for (i = 1; i <= cnt; ++i) 34 for (j = p[i]; j <= n; j += p[i]) 35 for (t = j; t % p[i] == 0; t /= p[i]) ++c[i]; 36 for (i = 1; i <= cnt; ++i) 37 (ans *= (c[i] << 1 | 1)) %= mod; 38 printf("%lld ", ans); 39 return 0; 40 }