Bzoj-2301 [HAOI2011]Problem b 容斥原理,Mobius反演,分块
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2301
题意：多次询问，求有多少对数满足 gcd(x,y)=k, a<=x<=b, c<=y<=d。
对于有下界的区间，容易想到用容斥原理做。然后如果直接用Mobius反演定理做，那么每次询问的复杂度是O(n/k)，如果k=1的话，那么总体就是O(n^2)的复杂度了，会TLE。这样用到了分快优化，注意到 n/i ,在连续的k区间内存在，n/i=n/(i+k)，因此能用分块优化。由于n/d，最多有2*sqrt(n)不相同的数，因此每次询问复杂度2*sqrt(n)+2*sqrt(m)..
详细内容推荐看：<POI XIV Stage.1 Queries Zap>
1 //STATUS:C++_AC_2052MS_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 long long LL; 34 typedef unsigned long long ULL; 35 //const 36 const int N=50010; 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 int T,n,a,b,c,d,k; 59 int isprime[N],mu[N],prime[N],sum[N]; 60 int cnt; 61 void Mobius(int n) 62 { 63 int i,j; 64 //Init phi[N],prime[N],全局变量初始为0 65 cnt=0;mu[1]=1; 66 for(i=2;i<=n;i++){ 67 if(!isprime[i]){ 68 prime[cnt++]=i; //prime[i]=1;为素数表 69 mu[i]=-1; 70 } 71 for(j=0;j<cnt && i*prime[j]<=n;j++){ 72 isprime[i*prime[j]]=1; 73 if(i%prime[j]) 74 mu[i*prime[j]]=-mu[i]; 75 else {mu[i*prime[j]]=0;break;} 76 } 77 } 78 } 79 80 LL solve(int n,int m) 81 { 82 int i,j,la; 83 LL ret=0; 84 if(n>m)swap(n,m); 85 for(i=1,la=0;i<=n;i=la+1){ 86 la=Min(n/(n/i),m/(m/i)); 87 ret+=(LL)(sum[la]-sum[i-1])*(n/i)*(m/i); 88 } 89 return ret; 90 } 91 92 int main(){ 93 // freopen("in.txt","r",stdin); 94 int i,j; 95 LL ans; 96 Mobius(50000); 97 for(i=1;i<50000;i++)sum[i]=sum[i-1]+mu[i]; 98 scanf("%d",&T); 99 while(T--) 100 { 101 scanf("%d%d%d%d%d",&a,&b,&c,&d,&k); 102 ans=solve(b/k,d/k)-solve((a-1)/k,d/k) 103 -solve((c-1)/k,b/k)+solve((a-1)/k,(c-1)/k); 104 printf("%lld\n",ans); 105 } 106 return 0; 107 }
