/* ***********************************************
Author :
Created Time :2015/5/26 8:02:37
File Name :1.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 1<<30
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
return a>b;
}
int gcd(int a,int b){
return a==0?b:(gcd(b%a,a));
}
int main()
{
#ifndef ONLINE_JUDGE
//freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
ll y0,x0,ans;
ll y;
while(scanf("%I64d%I64d",&x0,&y0)!=EOF){
ans=0;
for(int i=x0;i<=sqrt(x0*y0);i++){
if(y0%i==0&&i%x0==0){
y=x0*y0/i;
if(gcd(i,y)==x0)ans++;
}
}
printf("%I64d\n",ans);
}
return 0;
}

#include <iostream>

using namespace std;

long int commonDivisor(long int a, long int b){
int max,min;
if(a > b){
max = a;
min = b;
} else {
max = b;
min = a;
}
long int temp;
while(max % min != 0){
temp = max % min;
max = min;
min = temp;
}
return min;
}

long int commonMultiple(long int a, long int b){
long int div= commonDivisor(a, b);
return (a/div)*(b/div)*div;
}

int main(){
long int div;
long int mul;
cin >> div >> mul;
if((div > mul) || (mul%div != 0)){
cout << 0 << endl;
return 0;
}
int upBound = mul / div;
int mount = 0;
for(int x = 1; x <= upBound; x++)
for(int y = 1; y <= upBound; y++)
if(commonDivisor(div * x, div * y) == div && commonMultiple(div * x, div * y) == mul)
mount++;
cout << mount << endl;
}

program zhen;
var
i:longint;
x,y,ans,t,a:int64;
function gcd(a,b:int64):int64;
begin
if b=0 then exit(a) else exit(gcd(b,a mod b));
end;
procedure print(num:int64);
begin
writeln(num);
halt;
end;
begin
readln(x,y);
if x=y then print(1);
a:=x*y;
ans:=0;
for i:=1 to trunc(sqrt(a)) do
if a mod i=0 then
begin
t:=gcd(i,a div i);
if (t=x) and (a div t=y) then inc(ans);
end;
print(ans*2);
end.
简单数学知识，a*b=lcm(a,b)*gcd(a,b),既AB的积等于他们的最大公约数和最小公倍数之积，然后枚举下就可以了

渣渣也来水一发，沾沾楼下各位大神的仙气！

编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 548 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 552 KiB, score = 10
Accepted, time = 0 ms, mem = 556 KiB, score = 50

代码
#include <iostream>
#include<cmath>
using namespace std;
int gcd(int m, int n)
{
while (m != n)
{
if (m>n) m -= n;
else n -= m;
}
return m;
}

int main()
{
long long x, y, hahaha = 0, hahahaha = 0, ha, haha;
cin >> x >> y;
ha = x * y;
hahahaha = sqrt(double(y/x));
for (int i = 1; i <= hahahaha; i++)
{
haha=i*x;
if (y % haha == 0)
if(gcd(haha,ha/haha)==x)
hahaha++;
}

cout << 2 * hahaha << endl;
return 0;
}

//分析：
//要点：P*Q = x0*y0,即两个数的最大公约数和最小公倍数之积等于这两个数的积，然后根据这一点分析符合要求的数。我们先求出x0*y0，然后枚举P从i=2到i=根号x0*y0，只要满足x0*y0是i的倍数，且P,Q均为x0的倍数，且y0为x0，y0的倍数，并且还要满足P，Q的最小公约数是x0，因为15，12和12，15都满足，所以最后将答案*2即可。
#include<iostream>
#include<math.h>
using namespace std;

int x, y;

void swap(int &a, int &b) {//交换a，b
    int temp = a;
    a = b;
    b = temp;
}
int gcd(int a, int b) {//求a，b的最大公约数
    if (a < b) swap(a, b);
    int r = a%b;
    while (r) {
        a = b;
        b = r;
        r = a%b;
    }
    return b;
}
int main() {
    cin >> x >> y;
    long long a = x*y;
    int ans = 0;
    for (int i = 2; i <= sqrt(a)+1; i++) {
        if (a%i == 0 && i%x == 0 && y%i == 0 && (a / i) % x == 0 && y % (a / i) == 0) {
            int m = i, n = a / i;
            if (gcd(n, m) == x) ans++;
        }
    }
    cout << ans*2 << endl;
    return 0;
}

终于遇到一个不硬的柿子 T_T

#include <stdio.h>

int zuida(long x,long y,long z)
{
long temp;
if( x%z!=0 || y%z!=0 ) return 0;
while(x!=y)
{
if(x>y) {temp=x;x=y;y=temp;}
y=y-x;
}
if(y!=z) return 0;
else return 1;
}

int main()
{
long i,j,a,b,sum=0;
scanf("%ld %ld",&a,&b);

for(i=a;i<=b;i++)
for(j=a;j<=b;j++)
{
if(i*j != a*b) continue;
if( zuida(i,j,a) )
sum++;
}
printf("%ld",sum);
return 0;
}

import re
import sys

def readln():
    return map(int, input().split())

def read():
    s = input()
    k = []
    r = []
    k = re.split(' ',s)
    for i in k:
        r.append(int(i))
    return r

def write(e,ed):
    for i in e:
        print(i,end=ed)
    print()

def gcd(a,b):
    if b == 0:
        return a
    else:
        return gcd(b,a%b)

a,b = readln()
if a<b:
    a,b = b,a
f = gcd(a,b)
if f != b:
    print(0)
else:
    a //= f
    ans = 0
    for i in range(1,a+1):
        if a % i == 0:
            if a//i<i:
                r=i
                t=a//i
            else:
                r=a//i
                t=i
            
            if gcd(r,t) == 1:
                ans+=1
                #print(i*f,a//i*f)
    print(ans)

啦啦啦

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <cstdlib>
using namespace std;

int n,m;

int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}

int main()
{
    int ans=0;
    cin>>n>>m;
    int a=n*m;
    for(int i=2;i<=a;i++)
    {
        if(a%i!=0)
            continue;
        else
        {
            int j=a/i;
            if(gcd(i,j)==n)
                ans++;
        }
    }
    cout<<ans<<endl;
    return 0;
}
     

这题主要是对于p，q两个数的查找。
（定义：x0=x1; y0=y1）
最先想到的是把每种可能全找出来，统计结果。
如果我们需要找p，q两个数，为了优化时间，只需从1..x2/x1查找即可（也就是同除p，q两个数的最大公约数）。
因为 m最坏情况是1000000，可以在一秒内通过
所以 线性查找可以解决。

几个需要注意的点：
1.查找是***必须***先判断x2 mod x1（也就是c++中的x2％x1）的结果是否不为0，如不为0，则直接输出0，即没有符合条件的x2,x1；
2.需要判断x2,x1是否有公约数。

人民喜闻乐见的代码：

uses math;
var x1,x2,p,i,j,ans:longint;

function pd(x,y:longint):byte;
var m,i:longint;
    begin
        m:=min(x,y);
        pd:=0;
        for i:=2 to m do
            if (x mod i=0) and (y mod i=0) then begin pd:=1; exit; end;
    end;

    begin
        readln(x1,x2);
        if x2 mod x1<>0 then begin write(0); exit; end else
            begin
                p:=x2 div x1;
                i:=1;
                while i<=p do
                    begin
                        if p mod i=0 then 
                            begin 
                                j:=p div i; 
                                if pd(i,j)=0 then inc(ans);
                            end;
                        inc(i);
                    end;
            end;
        write(ans);
    end.                    

ps 欢迎指教

int gcd(int a, int b)
{
    if(a < b)
    {
        swap(a, b);   
    }

    if(b == 0)
    {
        return a;
    }

    if((a & 0x1) == 0 && (b & 0x1) == 0) 
    {
        return 2 * gcd(a >> 1, b >> 1);
    }

    if((a & 0x1) == 0 && (b & 0x1) != 0) 
    {
        return gcd(a >> 1, b);
    }

    if((a & 0x1) != 0 && (b & 0x1) == 0) 
    {
        return gcd(a, b >> 1);
    }

    if((a & 0x1) != 0 && (b & 0x1) != 0) 
    {
        return gcd((a - b) >> 1, b);
    }
}

二进制版gcd，加不加都是0ms

数据真是高端大气上档次。。
c++
#include <cstdio>
int n,m,x,p,q,ans=0,t=2;
int gcd(int x,int y)
{
if (y==0) return x;
else return gcd(y,x%y);
}
int main()
{
scanf("%d%d",&n,&m);
p=n*m;q=n;
while(q<=m)
{
x=p/q;
if(x*q==p)
if(gcd(x,q)==n) ans++;
q=n*t++;
}
printf("%d",ans);
}

这样的程序都能0ms，也是醉了

#include <stdio.h>
#include <stdlib.h>

int corr(int x0,int y0,int p,int q)
{
    int i=1,j;
    j=i*p;
    while(j%q!=0)
    {
        i++;
        j=i*p;
    }
    if(j!=y0)
        return 0;
    return 1;
}
int main()
{
    int x0,y0,p,q,sum=0,ji;
    scanf("%d%d",&x0,&y0);
    ji=x0*y0;
    for(p=x0;p<=y0;p++)
    {
        q=ji/p;
        if(corr(x0,y0,p,q))
            sum++;
    }
    printf("%d",sum);
    return 0;
}

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

int n,m,a,b,k;

int gcd(int a,int b)
{
    return b==0 ? a:gcd(b,a%b);
}

int main()
{
    cin>>n>>m;
    k=n*m;
    a=n;
    int ans=0;
    int kk=2;
    while(a<=m)
    {
        b=k/a;
        if(a*b==k)
            if(gcd(a,b)==n)
            {
                ans++;
                //cout<<a<<endl;
            }
        a=n*kk;
        kk++;
    }
    cout<<ans;
    return 0;
}