洛谷 P3455 [POI2007]ZAP-Queries || 洛谷P2522,bzoj2301

https://www.luogu.org/problemnew/show/P3455

就是https://www.cnblogs.com/hehe54321/p/9315244.html里面的方法2了,升级版的整除分块,可以两个变量一起搞

预处理莫比乌斯函数的前缀和之后就可以每次$O(\sqrt{n}+\sqrt{m})$回答

那篇题解里面用了一个技巧:${\lfloor}\frac{{\lfloor}\frac{a}{b}{\rfloor}}{c}{\rfloor}={\lfloor}\frac{a}{bc}{\rfloor}$

(当然a,b,c都为正整数)

证了好久。。。

这么证:

设${\lfloor}\frac{a}{bc}{\rfloor}=p$,则p为整数,且$p<=\frac{a}{bc}<p+1$

则$pc<=\frac{a}{b}<pc+c$

而$pc$与$pc+c$都为整数

因此$pc<={\lfloor}\frac{a}{b}{\rfloor}<pc+c$

所以$p<=\frac{{\lfloor}\frac{a}{b}{\rfloor}}{c}<p+1$

所以${\lfloor}\frac{{\lfloor}\frac{a}{b}{\rfloor}}{c}{\rfloor}=p={\lfloor}\frac{a}{bc}{\rfloor}$

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<vector>
 5 using namespace std;
 6 #define fi first
 7 #define se second
 8 #define mp make_pair
 9 #define pb push_back
10 typedef long long ll;
11 typedef unsigned long long ull;
12 typedef pair<int,int> pii;
13 #define N 50100
14 ll prime[N+100],len,mu[N+100],dd[N+100];
15 bool nprime[N+100];
16 ll a,c,n,m,k,ans,ed;
17 int main()
18 {
19     ll i,j,T,TT;
20     mu[1]=1;
21     for(i=2;i<=N;i++)
22     {
23         if(!nprime[i])    prime[++len]=i,mu[i]=-1;
24         for(j=1;j<=len&&i*prime[j]<=N;j++)
25         {
26             nprime[i*prime[j]]=1;
27             if(i%prime[j]==0)    {mu[i*prime[j]]=0;break;}
28             else    mu[i*prime[j]]=-mu[i];
29         }
30     }
31     for(i=1;i<=N;i++)    dd[i]=dd[i-1]+mu[i];
32     scanf("%lld",&T);
33     for(TT=1;TT<=T;TT++)
34     {
35         scanf("%lld%lld%lld",&n,&m,&k);n/=k;m/=k;
36         ans=0;
37         if(n>m)    swap(n,m);
38         for(i=1;i<=n;i=j+1)
39         {
40             j=min(n,min(n/(n/i),m/(m/i)));
41             ans+=(dd[j]-dd[i-1])*(n/i)*(m/i);
42         }
43         printf("%lld\n",ans);
44     }
45     return 0;
46 }

https://www.luogu.org/problemnew/show/P2522

https://www.lydsy.com/JudgeOnline/problem.php?id=2301

这题基本一样的,就是加个容斥。。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<vector>
 5 using namespace std;
 6 #define fi first
 7 #define se second
 8 #define mp make_pair
 9 #define pb push_back
10 typedef long long ll;
11 typedef unsigned long long ull;
12 typedef pair<int,int> pii;
13 #define N 50100
14 ll prime[N+100],len,mu[N+100],dd[N+100];
15 bool nprime[N+100];
16 ll a,c,n,m,k;
17 ll calc(ll n,ll m)
18 {
19     if(n==0||m==0)    return 0;
20     ll ans=0;
21     if(n>m)    swap(n,m);
22     n/=k;m/=k;
23     for(ll i=1,j;i<=n;i=j+1)
24     {
25         j=min(n,min(n/(n/i),m/(m/i)));
26         ans+=(dd[j]-dd[i-1])*(n/i)*(m/i);
27     }
28     return ans;
29 }
30 int main()
31 {
32     ll i,j,T,TT;
33     mu[1]=1;
34     for(i=2;i<=N;i++)
35     {
36         if(!nprime[i])    prime[++len]=i,mu[i]=-1;
37         for(j=1;j<=len&&i*prime[j]<=N;j++)
38         {
39             nprime[i*prime[j]]=1;
40             if(i%prime[j]==0)    {mu[i*prime[j]]=0;break;}
41             else    mu[i*prime[j]]=-mu[i];
42         }
43     }
44     for(i=1;i<=N;i++)    dd[i]=dd[i-1]+mu[i];
45     scanf("%lld",&T);
46     for(TT=1;TT<=T;TT++)
47     {
48         scanf("%lld%lld%lld%lld%lld",&a,&n,&c,&m,&k);
49         printf("%lld\n",calc(n,m)-calc(a-1,m)-calc(n,c-1)+calc(a-1,c-1));
50     }
51     return 0;
52 }

 

posted @ 2018-07-16 16:01  hehe_54321  阅读(...)  评论(...编辑  收藏
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