Bzoj 1101: [Poi2007]zap

1101: [Poi2007]zap time limit:10 Sec Memory limit:162 MB
submit:2262 solved:895
[Submit] [Status] [Discuss] Description

FGD is cracking a cipher, and he needs to answer a number of similar questions: for a given integer, a, B, and D, how many positive integers to X, Y, meet x<=a
, Y<=b, and gcd (x, y) =d. As a classmate of FGD, FGD hopes to get your help.

Input

The first line contains a positive integer n, which indicates that there are altogether n groups of queries. (1<=n<= 50000) next n lines, each line represents one query, three per line
positive integers, respectively, for A,b,d. (1<=d<=a,b<=50000)

Output

For each set of queries, output to the output file zap.out a positive integer that represents the integer logarithm that satisfies the condition.

Sample Input2
4 5 2
6 4 3Sample Output3
2
For the first set of queries, an integer pair that satisfies the condition has (2,2), (2,4), (4,2). For the second set of queries, an integer pair that satisfies the condition has (
6,3), (3,3). HINT Source Analysis:

In fact, the equivalent of seeking n/d,m/d limit, gcd (i,j) ==1 the number of IJ

The same practice as http://www.cnblogs.com/neighthorn/p/6214769.html ...

Note that the question O (n) is definitely not going to be ... Without looking at the data range, I'm going to go to the tle.

Because (n/i) * (m/i) The value of up to sqrt (n) +sqrt (m), so we pre-processing μ prefix and, segmentation calculation is good ...

Code:

1#include <algorithm>2#include <iostream>3#include <cstring>4#include <cstdio>5 //by Neighthorn6 using namespacestd;7 //Dapeng Day with the wind, soaring 90,000 miles8 9 Const intmaxn=50000+5;Ten  One intN,M,D,CAS,CNT,MIU[MAXN],VIS[MAXN],PRIME[MAXN]; A  - Long Longans=0; -  theSigned Main (void){ -memset (Vis,0,sizeof(VIS)); Cnt=0; miu[1]=1; -      for(intI=2; i<=50000; i++){ -         if(!Vis[i]) +prime[++cnt]=i,vis[i]=1, miu[i]=-1; -          for(intj=1; j<=cnt&&prime[j]*i<=50000; j + +){ +vis[i*prime[j]]=1; A             if(i%prime[j]==0){ atmiu[i*prime[j]]=0; Break; -             } -miu[i*prime[j]]=-Miu[i]; -         } -     } -      for(intI=1; i<=50000; i++) inmiu[i]+=miu[i-1]; -scanf"%d",&CAs); to      while(cas--){ +scanf"%d%d%d",&n,&m,&d); -n/=d,m/=d;ans=0; the         if(n>m) * swap (n,m); $          for(intI=1, r;i<=n;i=r+1){Panax NotoginsengR=min (n/(n/i), m/(m/i)); -ans+= (Long Long) (miu[r]-miu[i-1]) * (n/i) * (m/i); the         } +printf"%lld\n", ans); A     } the     return 0; +}

View Code

by Neighthorn

Bzoj 1101: [Poi2007]zap