BZOJ 2301 Problem B(莫比乌斯反演)

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

题意:给a,b,c,d,k,求gcd(x,y)==k的个数(a<=x<=b,c<=y<=d)

思路:假设F(a,b)代表gcd(x,y)==k 的个数(1<=x<=a,1<=y<=b)

那么这是满足区间加减的

ans=F(b,d)-F(b,c)-F(a,d)+F(a,c)

剩下的就和Zap一样了

 1 #include<algorithm>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<iostream>
 6 int mul[200005],p[200005],mark[200005],sum[200005];
 7 int read(){
 8     char ch=getchar();int t=0,f=1;
 9     while (ch<'0'||ch>'9'){if (ch=='0') f=-1;ch=getchar();}
10     while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
11     return t*f;
12 }
13 void init(){
14     mul[1]=1;
15     for (int i=2;i<=50000;i++){
16         if (!mark[i]){
17             p[++p[0]]=i;
18             mul[i]=-1;
19         }
20         for (int j=1;j<=p[0]&&p[j]*i<=50000;j++){
21             mark[i*p[j]]=1;
22             if (i%p[j]) mul[p[j]*i]=mul[i]*(-1);
23             else{
24                 mul[p[j]*i]=0;
25                 break;
26             } 
27         }
28     }
29     sum[0]=0;
30     for (int i=1;i<=50000;i++) sum[i]=sum[i-1]+mul[i];
31 }
32 int cal(int a,int b){
33     if (a>b) std::swap(a,b);
34     int res=0;
35     for (int i=1,j;i<=a;i=j+1){
36         j=std::min(a/(a/i),b/(b/i));
37         res+=(a/i)*(b/i)*(sum[j]-sum[i-1]);
38     }
39     return res;
40 }
41 int main(){
42     int T=read();
43     init();
44     while (T--){
45         int a=read(),b=read(),c=read(),d=read(),k=read();
46         a--;c--;
47         printf("%d\n",std::max(0,cal(b/k,d/k)+cal(a/k,c/k)-cal(b/k,c/k)-cal(a/k,d/k)));
48     }
49 }

 

posted @ 2016-06-14 11:19  GFY  阅读(...)  评论(...编辑  收藏