NOIP 模擬 $26\; \rm 神炎皇$
阿新 • • 發佈:2021-07-31
題解
題解 \(by\;zj\varphi\)
一道 \(\varphi()\) 的題。
對於一個合法的數對,設它為 \((a*m,b*m)\) 則 \(((a+b)*m)|a*b*m^2\),所以 \((a+b)|a*b\),因為 \(\gcd(a,b)=1\),所以 \(a+b|m\)
那麼設 \(a+b=k\),且 \(k*m\le n\),那麼 \(k\) 最大為 \(\sqrt n\),所以列舉 \(k\) 即可
Code:
#include<bits/stdc++.h> #define ri register signed #define p(i) ++i using namespace std; namespace IO{ char buf[1<<21],*p1=buf,*p2=buf,OPUT[100]; #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++ template<typename T>inline void read(T &x) { ri f=1;x=0;register char ch=gc(); while(!isdigit(ch)) {if (ch=='-') f=0;ch=gc();} while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=gc();} x=f?x:-x; } template<typename T>inline void print(T x) { if (x<0) putchar('-'),x=-x; if (!x) return putchar('0'),(void)putchar('\n'); ri cnt(0); while(x) OPUT[p(cnt)]=x%10,x/=10; for (ri i(cnt);i;--i) putchar(OPUT[i]^48); return (void)putchar('\n'); } } using IO::read;using IO::print; namespace nanfeng{ #define FI FILE *IN #define FO FILE *OUT template<typename T>inline T cmax(T x,T y) {return x>y?x:y;} template<typename T>inline T cmin(T x,T y) {return x>y?y:x;} typedef long long ll; static const int N=1e7+7; int vis[N],phi[N],prim[N],nm[N],cnt,sn; ll n,ans; inline void Getphi(int n) { for (ri i(2);i<=n;p(i)) { if (!vis[i]) phi[i]=i-1,vis[prim[p(cnt)]=i]=i; for (ri j(1);j<=cnt&&prim[j]*i<=n;p(j)) { int nw=i*prim[j]; vis[nw]=prim[j]; if (vis[i]==prim[j]) { phi[nw]=phi[i]*prim[j]; break; } else phi[nw]=phi[i]*(prim[j]-1); } } } inline int main() { //FI=freopen("nanfeng.in","r",stdin); //FO=freopen("nf.out","w",stdout); read(n); sn=sqrt(n); Getphi(sn); for (ri i(2);i<=sn;p(i)) ans+=(ll)phi[i]*(n/i/i); print(ans); return 0; } } int main() {return nanfeng::main();}