Bzoj-2005 能量采集 gcd,递推

时间：2021-12-19 08:16:00

　　题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2005

　　题意：题目转换后的模型就是求Σ(gcd(x,y)), 1<=x<=n, 1<=y<=m。。

　　容易想到n^2logn的方法，ΣΣ(gcd(x,y)*2-1)，但是这里会超时，因此我们需要优化。我们令f[d]表示(x,y),1<=x<=n, 1<=y<=m的所有对数中gcd(x,y)=d的个数，那么容易求出所有对数中(x,y)的约数为d的个数为(n/d)*(m/d)，然后减去f[i*d],i>=2就行了...

 //STATUS:C++_AC_16MS_2052KB

 #include <functional>

 #include <algorithm>

 #include <iostream>

 //#include <ext/rope>

 #include <fstream>

 #include <sstream>

 #include <iomanip>

 #include <numeric>

 #include <cstring>

 #include <cassert>

 #include <cstdio>

 #include <string>

 #include <vector>

 #include <bitset>

 #include <queue>

 #include <stack>

 #include <cmath>

 #include <ctime>

 #include <list>

 #include <set>

 #include <map>

 using namespace std;

 //#pragma comment(linker,"/STACK:102400000,102400000")

 //using namespace __gnu_cxx;

 //define

 #define pii pair<int,int>

 #define mem(a,b) memset(a,b,sizeof(a))

 #define lson l,mid,rt<<1

 #define rson mid+1,r,rt<<1|1

 #define PI acos(-1.0)

 //typedef

 typedef __int64 LL;

 typedef unsigned __int64 ULL;

 //const

 const int N=;

 const int INF=0x3f3f3f3f;

 const int MOD=,STA=;

 const LL LNF=1LL<<;

 const double EPS=1e-;

 const double OO=1e15;

 const int dx[]={-,,,};

 const int dy[]={,,,-};

 const int day[]={,,,,,,,,,,,,};

 //Daily Use ...

 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 //End

 LL f[N];

 int n,m;

 int main(){

     freopen("in.txt","r",stdin);

     int i,j,low;

     LL ans;

     scanf("%d%d",&n,&m);

     low=Min(n,m);

     ans=;

     for(i=low;i>;i--){

         f[i]=(LL)(n/i)*(m/i);

         for(j=i+i;j<=low;j+=i)f[i]-=f[j];

         ans+=f[i]*(i*-);

     }

     printf("%lld\n",ans);

     return ;

 }

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