# bzoj2721 [Violet 5] cherry blossoms

Source: Internet
Author: User

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Output

Sample Input Sample Output HINT

The formula problem of the egg ache ... According to test instructions 1/x+1/y=1/z, make y=z+d, and then

1/x+1/(z+d) =1/z

(x+z+d)/(XZ+XD) =1/z

Xz+z^2+dz=xz+xd

Z^2+dz=xd

X=z^2/d+z

Obviously x is a positive integer depending on the D can divide z^2

Each d corresponds to a unique x, so the answer is the approximate number of z^2.

(n!) The approximate number of ^2

`#include <cstdio> #include <iostream> #include <cstring> #include <cstdlib> #include < algorithm> #include <cmath> #include <queue> #include <deque> #include <set> #include <map > #include <ctime> #define LL long long#define INF 0x7ffffff#define pa pair<int,int> #define MOD 1000000007#    Define MX 1000000using namespace Std;inline ll read () {ll 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;}    inline void Write (LL a) {if (a<0) {printf ("-"); a=-a;}    if (a>=10) write (A/10); Putchar (a%10+ ' 0 ');} inline void Writeln (LL a) {write (a);p rintf ("\ n");} BOOL PRIME[MX+10]; LL rep[mx+10];int N;    LL ans=1;inline void Shai () {memset (prime,1,sizeof (prime));    prime[1]=0;        for (int i=2;i<=n;i++) if (Prime[i]) {LL res=i;            while (LL) n/res>0 {rep[i]+= (ll) n/res;  Res*=i;      } for (int j=2*i;j<=n;j+=i) prime[j]=0;    }}int Main () {n=read (); Shai ();        if (n==1) {printf ("1\n");    return 0;        } for (int i=1;i<=n;i++) if (Prime[i]) {ans=ans* (2*rep[i]+1)%mod; } printf ("%lld\n", ans);}`

bzoj2721 [Violet 5] cherry blossoms

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