CSU 1325: a very hard problem

1325: A very hard problemtime limit: 3 sec memory limit: 160 MB
Submit: 203 solved: 53
[Submit] [Status] [web board] Description

CX old wet often gets hacked, much hacked, and I am also numb. As a result, he often heard someone say "Animals" with deep affection!

One day, CX experienced a sudden whim, which gave everyone a difficult problem and claimed that he could continue to attack him if he could answer the question accurately. Otherwise, he would have to fight back.

This problem is:

Two numbers p and q are given, and then two numbers A and B are given for each question. Please find them separately:

1. How many sequential number pairs (x, y) meet the requirements of 1 <= x <= A, 1 <= Y <= B, and gcd (x, y) it is an approximate number of P;

2. How many sequential number pairs (x, y) meet the requirements of 1 <= x <= A, 1 <= Y <= B, and gcd (x, y) is a multiple of P.

 

 

Input

 

There is only one set of test data.

The first row has two numbers: p and q. (1 <p <10 ^ 7, 1 <q <1000 .)

Next, there will be Q rows, with two numbers A and B in each row. (1 <a, B <10 ^ 7)

 

 

 

Output

 

Output A total of Q rows. There are two numbers in each row. Separated by spaces.

The two corresponding answers in the question description.

 

(X, y) = (2, 3) and (x, y) = (3, 2) are considered as two different sequential number pairs!

 

 

 

Sample Input

6 38 815 3213 77

Sample output

58 1423 10883 24

Hint

 

For 64-bit integer types, use LLD, or CIN, cout. T_t

Csu_scsi

This is the question of last year's competition. At that time, I felt it was difficult to implement it with Euler.

Later, I learned to use the Mobius Inversion, but it has timed out. Whether the block is used or time-out.

Question: omitted

Idea: For the second type, direct (A/P) * (B/P) is the answer. It is not hard to understand.

For the first case:

Set

G (p) represents each factor of enumeration P. Di is used to calculate GCD (x, y) = di (1 <= x <= A, 1 <= Y <= B).

It is the value required by the question.

If we enumerate the divisor of each P at this time, it will time out.

This sub-statement can be converted to another t = di * X, then the sub-statement can be converted:

In this case, we only need to pre-process the last part, and then we can solve this problem with SQRT (min (a, B) time.

How to pre-process this part?

First, create a table to obtain the U [].

In this formula, Tom (t) = sigma (U [di], T % di = 0 & Di is a factor of P );

Since P is unique, We can find its factor, and then use its factor to filter the array hxl [], which is Tom (t );

The last hxl [], the first n sums, which are done in parts.

 1 #include<iostream>  2 #include<stdio.h>  3 #include<cstring>  4 #include<cstdlib>  5 #include<math.h>  6 using namespace std;  7    8 typedef long long LL;  9 const int maxn = 1e7+1; 10 bool s[maxn]; 11 int prime[670000],len = 0; 12 int yz[10002],ylen; 13 int mu[maxn]; 14 int hxl[maxn]; 15 void  init() 16 { 17     memset(s,true,sizeof(s)); 18     mu[1] = 1; 19     for(int i=2;i<maxn;i++) 20     { 21         if(s[i] == true) 22         { 23             prime[++len]  = i; 24             mu[i] = -1; 25         } 26         for(int j=1;j<=len && ((long long)prime[j])*i<maxn;j++) 27         { 28             s[i*prime[j]] = false; 29             if(i%prime[j]!=0) 30                 mu[i*prime[j]] = -mu[i]; 31             else32             { 33                 mu[i*prime[j]] = 0; 34                 break; 35             } 36         } 37     } 38 } 39 void solve(int  p) 40 { 41     ylen = 0; 42     int k = (int)sqrt(p*1.0); 43     int tmp; 44     for(int i=1;i<=k;i++) 45     { 46         if(p%i==0) 47         { 48             yz[++ylen] = i; 49             tmp = p/i; 50             if(tmp!=i) yz[++ylen] = tmp; 51         } 52     } 53     for(int i=1;i<=ylen;i++) 54     { 55         for(int j=yz[i],k=1;j<maxn;j=j+yz[i],k++) 56             hxl[j]=hxl[j]+mu[k]; 57     } 58     for(int i=2;i<maxn;i++) hxl[i] = hxl[i]+hxl[i-1]; 59 } 60 int main() 61 { 62     init(); 63     int  p,q,A,B; 64     scanf("%d%d",&p,&q); 65     solve(p); 66     while(q--) 67     { 68         scanf("%d%d",&A,&B); 69         if(A>B) swap(A,B); 70         long long ans1 = 0,ans2 = 0; 71         for(int i=1,la = 0;i<=A; i=la+1) 72         { 73             la = min(A/(A/i),B/(B/i)); 74             ans1 = ans1+(long long)(hxl[la]-hxl[i-1])*(A/i)*(B/i); 75         } 76         ans2 = (A/p)*(B/p); 77         printf("%lld %lld\n",ans1,ans2); 78     } 79     return 0; 80 } 81   82 /************************************************************** 83     Problem: 1325 84     User: 987690183 85     Language: C++ 86     Result: Accepted 87     Time:1052 ms 88     Memory:92044 kb 89 ****************************************************************/

 

CSU 1325: a very hard problem