标签：article blog 多少 row mod const .com 连续 review
题意：对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，
且gcd(x,y) = k，gcd(x,y)函数为x和y的最大公约数。
1≤n≤50000，1≤a≤b≤50000，1≤c≤d≤50000，1≤k≤50000
思路：第一题反演……
利用容斥原理将原询问拆成4个，问题就转化为：
1<=i<=trunc(a div k)，1<=j<=trunc(b div k)，gcd(i,j)=1的(i,j)数对个数
令f(i)表示满足gcd(x,y)=i时(x,y)的对数，F(i)表示满足i|gcd(x,y)的(x,y)的对数
显然F(i)=trunc(n div i)*trunc(m div i)
f(1)=sigma u(d)*trunc(n div d)*trunc(m div d) (1<=d<=n)
观察可得trunc(n div d)*trunc(m div d)只有2根号n个取值，且每个取值对应的u(i)是连续的一段
然后就可以记录u的前缀和来优化
From http://m.blog.csdn.net/article/details?id=50590197
1 //uses sysutils; 2 const max=50000; 3 var mu,sum,prime:array[0..max]of longint; 4 flag:array[0..max]of longint; 5 a,b,c,d,k,i,j,t,m,cas,v:longint; 6 tmp:double; 7 8 procedure swap(var x,y:longint); 9 var t:int64; 10 begin 11 t:=x; x:=y; y:=t; 12 end; 13 14 function clac(n,m:longint):int64; 15 var i,t1,t2,pos:longint; 16 x,y:int64; 17 begin 18 if n>m then swap(n,m); 19 clac:=0; i:=1; 20 while i<=n do 21 begin 22 x:=n div i; y:=m div i; 23 t1:=n div x; 24 t2:=m div y; 25 if t1<t2 then pos:=t1 26 else pos:=t2; 27 clac:=clac+x*y*(sum[pos]-sum[i-1]); 28 i:=pos+1; 29 end; 30 end; 31 32 begin 33 assign(input,‘bzoj2301.in‘); reset(input); 34 assign(output,‘bzoj2301.out‘); rewrite(output); 35 // tmp:=now; 36 read(cas); 37 mu[1]:=1; 38 for i:=2 to max do 39 begin 40 if flag[i]=0 then 41 begin 42 inc(m); prime[m]:=i; 43 mu[i]:=-1; 44 end; 45 j:=1; 46 while (j<=m)and(prime[j]*i<=max) do 47 begin 48 t:=prime[j]*i; 49 flag[t]:=1; 50 if i mod prime[j]=0 then 51 begin 52 mu[t]:=0; 53 break; 54 end; 55 mu[t]:=-mu[i]; 56 inc(j); 57 end; 58 end; 59 for i:=1 to max do sum[i]:=sum[i-1]+mu[i]; 60 for v:=1 to cas do 61 begin 62 read(a,b,c,d,k); 63 dec(a); dec(c); 64 a:=a div k; b:=b div k; c:=c div k; d:=d div k; 65 writeln(clac(b,d)-clac(b,c)-clac(a,d)+clac(a,c)); 66 end; 67 //writeln((now-tmp)*86400:0:2); 68 close(input); 69 close(output); 70 end.
标签：article blog 多少 row mod const .com 连续 review
原文地址：http://www.cnblogs.com/myx12345/p/6666420.html
