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...
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; 24 //#pragma comment(linker,"/STACK:102400000,102400000") 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 }
