Bzoj-2005 能量采集 gcd,递推

　　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 ;
 }