# Bzoj-2005 能量采集 gcd,递推

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...

 1 //STATUS:C++_AC_16MS_2052KB
2 #include <functional>
3 #include <algorithm>
4 #include <iostream>
5 //#include <ext/rope>
6 #include <fstream>
7 #include <sstream>
8 #include <iomanip>
9 #include <numeric>
10 #include <cstring>
11 #include <cassert>
12 #include <cstdio>
13 #include <string>
14 #include <vector>
15 #include <bitset>
16 #include <queue>
17 #include <stack>
18 #include <cmath>
19 #include <ctime>
20 #include <list>
21 #include <set>
22 #include <map>
23 using namespace std;
25 //using namespace __gnu_cxx;
26 //define
27 #define pii pair<int,int>
28 #define mem(a,b) memset(a,b,sizeof(a))
29 #define lson l,mid,rt<<1
30 #define rson mid+1,r,rt<<1|1
31 #define PI acos(-1.0)
32 //typedef
33 typedef __int64 LL;
34 typedef unsigned __int64 ULL;
35 //const
36 const int N=100010;
37 const int INF=0x3f3f3f3f;
38 const int MOD=100000,STA=8000010;
39 const LL LNF=1LL<<60;
40 const double EPS=1e-8;
41 const double OO=1e15;
42 const int dx[4]={-1,0,1,0};
43 const int dy[4]={0,1,0,-1};
44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
45 //Daily Use ...
46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
56 //End
57
58 LL f[N];
59 int n,m;
60
61 int main(){
62     freopen("in.txt","r",stdin);
63     int i,j,low;
64     LL ans;
65     scanf("%d%d",&n,&m);
66     low=Min(n,m);
67     ans=0;
68     for(i=low;i>0;i--){
69         f[i]=(LL)(n/i)*(m/i);
70         for(j=i+i;j<=low;j+=i)f[i]-=f[j];
71         ans+=f[i]*(i*2-1);
72     }
73     printf("%lld\n",ans);
74     return 0;
75 }

posted @ 2013-08-19 15:36  zhsl  阅读(252)  评论(0编辑  收藏  举报