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先预处理莫比乌斯函数,再分块求

不会用公式编辑TAT,直接贴题解吧。。

从结论来看貌似也能用容斥原理?

 1 #include<bits/stdc++.h>
 2 #define inc(i,l,r) for(int i=l;i<=r;i++)
 3 #define dec(i,l,r) for(int i=l;i>=r;i--)
 4 #define link(x) for(edge *j=h[x];j;j=j->next)
 5 #define mem(a) memset(a,0,sizeof(a))
 6 #define inf 1e9
 7 #define ll long long
 8 #define succ(x) (1<<x)
 9 #define NM 50000+5
10 using namespace std;
11 int read(){
12     int x=0,f=1;char ch=getchar();
13     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
14     while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
15     return x*f;
16 }
17 int mu[NM],sum[NM],a,b,c,d,k,T,p[NM],tot;
18 bool check[NM];
19 int slove(int x,int y){
20     int t,ans=0;
21     if((x/=k)>(y/=k))swap(x,y);
22     for(int i=1;i<=x;i=t+1){
23         t=min(x/(x/i),y/(y/i));
24         if(t>x)t=x;
25         ans+=(sum[t]-sum[i-1])*(x/i)*(y/i);
26     }
27     return ans;
28 }
29 int main(){
30     freopen("data.in","r",stdin);
31     T=read();
32     sum[1]=mu[1]=1;
33     inc(i,2,NM-5){
34         if(!check[i]){
35             p[++tot]=i;
36             mu[i]=-1;
37         }
38         sum[i]=sum[i-1]+mu[i];
39         inc(j,1,tot){
40             if(i*p[j]>NM-5)break;
41             check[i*p[j]]=true;
42             if(i%p[j])mu[i*p[j]]=-mu[i];
43             else break;
44         }
45     }
46 //    inc(i,1,10)printf("%d ",mu[i]);printf("\n");
47     while(T--){
48         a=read();b=read();c=read();d=read();k=read();
49         printf("%d\n",slove(b,d)-slove(a-1,d)-slove(b,c-1)+slove(a-1,c-1));
50     }
51     return 0;
52 }
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posted on 2016-02-03 13:33  onlyRP  阅读(...)  评论(...编辑  收藏