#include<iostream>
using namespace std;
int find(int a,int b)
{
int c;
if(a%b==0)
return b;
else
{
c=b;
b=a%b;
a=c;
find(a,b);

}
}
int main()
{
int x0,y0,a[10000]={0},s=1,t=1,n=0,tot=0,m,mm,nn;
cin>>x0>>y0;
if(x0>y0)
{
cout<<"0";
return 0;
}
m=x0*y0;
a[s]=x0;
while(a[s]<=y0)
{
if(y0%a[s]==0)
{
a[s]=x0*t;
s=s+1;
}
t=t+1;
a[s]=x0*t;
}
for(int i=1;i<=10000;++i)
{
if(a[i]!=0)
n=n+1;
if(a[i]==0)
break;
}
for(int i=1;i<n;++i)
{
for(int k=1;k<n;++k)
{
if(a[i]!=a[k])
{
mm=find(a[i],a[k]);
nn=a[i]*a[k]/mm;
if(mm==x0&&nn==y0)
tot=tot+1;
}
}
}
cout<<tot;
return 0;
}

var i,j,k,gcd,gbs,t,tot,zs:longint;
begin
readln(gcd,gbs);
if gbs mod gcd<>0 then begin
writeln(0);haltend;
t:=gbs div gcd;
i:=1;
while t>1 dobegin
inc(i);
if t modi=0 then
begin
inc(tot);
while tmod i=0 do
t:=t divi;
endend;

zs:=trunc(exp(ln(2)*tot));
writeln(zs);
end.

编译成功
测试数据 #0: Accepted, time = 0 ms, mem = 728 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 732 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 728 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 728 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 732 KiB, score = 10
Accepted, time = 0 ms, mem = 732 KiB, score = 50

代码有点丑。。。见谅！
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LOL;
long long int x,y,p,q,ans,temp;
LOL zuidagongyueshu(LOL a,LOL b)
{
if(a<b) swap(a,b);
if(b==0) return a;
else return zuidagongyueshu(b,a%b);
}
int main()
{
scanf("%d%d",&x,&y);
p=2;
q=3;
while(p<=y)
{
if((x*y)%p==0)
{
if((x*y)/p<=y)
{q=(x*y)/p;
temp=zuidagongyueshu(p,q);
if(temp==x&&p/temp*q==y) ans++;}
}
p++;
}
printf("%d",ans);
return 0;
}

编译成功

foo.pas(6,8) Warning: Variable "i" does not seem to be initialized
foo.pas(21,52) Warning: Variable "s" does not seem to be initialized

测试数据 #0: Accepted, time = 0 ms, mem = 820 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 816 KiB, score = 10

测试数据 #2: Accepted, time = 15 ms, mem = 816 KiB, score = 10

测试数据 #3: Accepted, time = 15 ms, mem = 816 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 820 KiB, score = 10

Accepted, time = 45 ms, mem = 820 KiB, score = 50

代码
var
x1,x2,i,j,s,y,r,a,b:int64;
begin
readln(x1,x2);
repeat
i:=i+x1;
j:=0;
repeat
j:=j+x1;
a:=j;
b:=i;
repeat
if a<b then begin r:=a; a:=b; b:=r; end; a:=a-b; until a=0; if (i*j div x1 =x2) and (b=x1 ) then s:=s+1; until j>=x2;
until i>=x2;
write(s);
end.

编译成功

foo.pas(6,8) Warning: Variable "i" does not seem to be initialized
foo.pas(21,52) Warning: Variable "s" does not seem to be initialized

测试数据 #0: Accepted, time = 0 ms, mem = 820 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 816 KiB, score = 10

测试数据 #2: Accepted, time = 15 ms, mem = 816 KiB, score = 10

测试数据 #3: Accepted, time = 15 ms, mem = 816 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 820 KiB, score = 10

Accepted, time = 45 ms, mem = 820 KiB, score = 50

代码
var
x1,x2,i,j,s,y,r,a,b:int64;
begin
readln(x1,x2);
repeat
i:=i+x1;
j:=0;
repeat
j:=j+x1;
a:=j;
b:=i;
repeat
if a<b then begin r:=a; a:=b; b:=r; end; a:=a-b; until a=0; if (i*j div x1 =x2) and (b=x1 ) then s:=s+1; until j>=x2;
until i>=x2;
write(s);
end.

编译成功

foo.pas(6,8) Warning: Variable "i" does not seem to be initialized
foo.pas(21,52) Warning: Variable "s" does not seem to be initialized

测试数据 #0: Accepted, time = 0 ms, mem = 820 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 816 KiB, score = 10

测试数据 #2: Accepted, time = 15 ms, mem = 816 KiB, score = 10

测试数据 #3: Accepted, time = 15 ms, mem = 816 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 820 KiB, score = 10

Accepted, time = 45 ms, mem = 820 KiB, score = 50

代码
var
x1,x2,i,j,s,y,r,a,b:int64;
begin
readln(x1,x2);
repeat
i:=i+x1;
j:=0;
repeat
j:=j+x1;
a:=j;
b:=i;
repeat
if a<b then begin r:=a; a:=b; b:=r; end; a:=a-b; until a=0; if (i*j div x1 =x2) and (b=x1 ) then s:=s+1; until j>=x2;
until i>=x2;
write(s);
end.

编译成功

foo.pas(6,8) Warning: Variable "i" does not seem to be initialized
foo.pas(21,52) Warning: Variable "s" does not seem to be initialized

测试数据 #0: Accepted, time = 0 ms, mem = 820 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 816 KiB, score = 10

测试数据 #2: Accepted, time = 15 ms, mem = 816 KiB, score = 10

测试数据 #3: Accepted, time = 15 ms, mem = 816 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 820 KiB, score = 10

Accepted, time = 45 ms, mem = 820 KiB, score = 50

代码
var
x1,x2,i,j,s,y,r,a,b:int64;
begin
readln(x1,x2);
repeat
i:=i+x1;
j:=0;
repeat
j:=j+x1;
a:=j;
b:=i;
repeat
if a<b then begin r:=a; a:=b; b:=r; end; a:=a-b; until a=0; if (i*j div x1 =x2) and (b=x1 ) then s:=s+1; until j>=x2;
until i>=x2;
write(s);
end.

编译成功

foo.pas(6,8) Warning: Variable "i" does not seem to be initialized
foo.pas(21,52) Warning: Variable "s" does not seem to be initialized

测试数据 #0: Accepted, time = 0 ms, mem = 820 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 816 KiB, score = 10

测试数据 #2: Accepted, time = 15 ms, mem = 816 KiB, score = 10

测试数据 #3: Accepted, time = 15 ms, mem = 816 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 820 KiB, score = 10

Accepted, time = 45 ms, mem = 820 KiB, score = 50

代码
var
x1,x2,i,j,s,y,r,a,b:int64;
begin
readln(x1,x2);
repeat
i:=i+x1;
j:=0;
repeat
j:=j+x1;
a:=j;
b:=i;
repeat
if a<b then
begin
r:=a;
a:=b;
b:=r;
end;
a:=a-b;
until a=0;
if (i*j div x1 =x2) and (b=x1 ) then s:=s+1;
until j>=x2;
until i>=x2;
write(s);
end.

其实就是杨辉三角一行的求和2^n 注意两数互质
var x0,y0,a:int64;
rec:array[1..100000]of integer;
ii,sum,i,n:longint;
procedure factor;
begin
n:=2;
ii:=0;
while n<=a do
begin
if a mod n=0 then
begin
while a mod n=0 do
a:=a div n;
inc(ii);
rec[ii]:=n;
end;
inc(n);
end;
end;

begin
read(x0,y0);
if y0 mod x0<>0 then begin write('0');halt;end;
a:=y0 div x0;
factor;
sum:=1;
for i:=1 to ii do
sum:=sum*2;
write(sum);
end.

测试数据 #0: Accepted, time = 15 ms, mem = 272 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 276 KiB, score = 10

测试数据 #2: Accepted, time = 15 ms, mem = 276 KiB, score = 10

测试数据 #3: Accepted, time = 15 ms, mem = 276 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 276 KiB, score = 10

#include<iostream>
using namespace std;
int a[10005];
/*思路：因为x0是最大公约数，y0是最小公倍数，所以，假设x,y 是一组满足条件的数，
那么有 x%x0 == 0 y%x0 == 0 x*y = x0*y0所以，我们先找到所有x0到
yo范围内的x0的倍数，这个倍数还要满足y0%x == 0 所以 我们用一个数组存
下这些数 比如 3- 60 就是 3 6 12 15 30 60 6个数 满足x%x0==0 y0%x==0
然后枚举这里面的每两个数的公约数,当然前提是x*y = x0*y0*/
int zzxc(int a,int b)//辗转相除法
{
int t;
while(b>0)
{
t=a%b;
a=b;
b=t;
}
return a;
}
int main()
{
int x,y,i,l=0,j,s=0;
cin>>x>>y;
for(i=x;i<=y;i+=x)
if(y%i==0)
a[l++]=i;
for(i=0;i<l;i++)
for(j=i+1;j<l;j++)
if(a[i]*a[j]==x*y&&zzxc(a[i],a[j])==x)
s+=2;
cout<<s<<endl;
return 0;
}

24K纯水啊。。。
#include <iostream>
#include <algorithm>
using namespace std;

long long gcd(long long a,long long b)
{
if(a<b) swap(a,b);
return (b==0)?a:gcd(b,a%b);
}
int main()
{
long long x0,y0; cin>>x0>>y0;
if(x0==y0) { cout<<1<<endl;return 0; }
long long t=x0*y0;
long long ans=0;
long long a,b,gab;
for(long long i=1;i*i<=t;++i)
{
if(!(t%i))
{
a=i;b=t/i;
// cout<<a<<" "<<b<<endl;
gab=gcd(a,b);
if(gab==x0 && a/gab*b==y0) ans++;
}
}

cout<<ans*2<<endl;
}

编译成功

Free Pascal Compiler version 2.6.4 [2014/03/06] for i386
Copyright (c) 1993-2014 by Florian Klaempfl and others
Target OS: Win32 for i386
Compiling foo.pas
foo.pas(22,47) Warning: Variable "ans" does not seem to be initialized
Linking foo.exe
27 lines compiled, 0.1 sec , 28176 bytes code, 1628 bytes data
1 warning(s) issued
测试数据 #0: Accepted, time = 0 ms, mem = 740 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 740 KiB, score = 10
测试数据 #2: Accepted, time = 125 ms, mem = 740 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 740 KiB, score = 10
测试数据 #4: Accepted, time = 109 ms, mem = 744 KiB, score = 10
Accepted, time = 234 ms, mem = 744 KiB, score = 50
var x,y,p,q,ans,s:longint;
function gcd(a,b:longint):longint;
begin
if b=0 then exit(a) else gcd:=gcd(b,a mod b);
end;

function acm(a,b,w:longint):longint;
begin
acm:=a*b div w;
end;

BEGIN
readln(x,y);
if y mod x<>0 then begin
writeln(0);
halt;
end else begin
for p:=x to y do
for q:=x to y do
begin
s:=gcd(p,q);
if (s=x) and (acm(p,q,s)=y) then inc(ans);
end;
writeln(ans);
end;
end.

#include <iostream>
using namespace std;
int n,m;
int main()
{
int i,j,k,l,q,w,e;
cin >>n>>m;
if ((n==0)||(m==0)||(m%n!=0)||(n>m)) {cout <<0;return 0;}
if (n==m) {cout <<1;return 0;}
m/=n;w=0;
for (i=1;i*i<=m;i++) if (m%i==0)
{
k=i;l=m/i;e=0;
for (j=2;j<=k;j++) if ((k%j==0)&&(l%j==0)) {e=1;break;}
if (e==1) continue;w+=2;
}
cout <<w<<endl;
}

内存开long long就莫名其妙的过了，说好的O(100000*1000000)呢？
测试数据 #0: Accepted, time = 15 ms, mem = 268 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #2: Accepted, time = 62 ms, mem = 268 KiB, score = 10
测试数据 #3: Accepted, time = 982 ms, mem = 272 KiB, score = 10
测试数据 #4: Accepted, time = 31 ms, mem = 268 KiB, score = 10
Accepted, time = 1090 ms, mem = 272 KiB, score = 50
代码
//暴力搜索
#include <iostream>

#include <stdio.h>
using namespace std;
long long x0,y0,ans;
int d,p,Q; //P与Q的最大公因数
inline void init(){
scanf("%d %d",&x0,&y0);
}
inline int gcd(int p,int q){
int s=1;
if(p<q)swap(p,q);
while(s!=0){s=p%q;p=q;;q=s;}//辗转相除法找出最大公因数
return p;
}
inline void dfs(){ //暴搜求出P,Q的个数
for(p=x0;p<=y0;p++) //使他们之积小于最小公倍数--剪枝
for(Q=x0;Q<=y0;Q++){
d=gcd(p,Q);
if(d==x0 && p*Q/d==y0)ans++; //判断是否符合条件
}
printf("%d\n",ans);
}
int main(){
init();
dfs();
return 0;
}

include<iostream>
using namespace std;
int gys(int x,int y){
if(x>y)swap(x,y);
if(x==0)return y;
else return gys(x,y%x);
}
int gbs(int x,int y){
if(x>y)swap(x,y);
int t=1;
while(y*t%x!=0)t++;
return t*y;
}
int main(){
int x,y,count=0,i;
long long s;
cin>>x>>y;
s=x*y;
for(i=x;i<=y;i++){
if(gys(i,s/i)==x && gbs(i,s/i)==y){
count++;
}
}
cout<<count<<endl;
return 0;
}

评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 464 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 464 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 464 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 472 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 464 KiB, score = 10
Accepted, time = 0 ms, mem = 472 KiB, score = 50
代码
#include<iostream>
using namespace std;

int gys(int x,int y){
if(x>y)swap(x,y);
if(x==0)return y;
else return gys(x,y%x);
}

int gbs(int x,int y){
if(x>y)swap(x,y);
int t=1;
while(y*t%x!=0)t++;
return t*y;
}

int main(){
int x,y,count=0,i;
long long s;
cin>>x>>y;
s=x*y;
for(i=x;i<=y;i++){
if(gys(i,s/i)==x && gbs(i,s/i)==y){
count++;
}
}
cout<<count<<endl;
return 0;
}

这道题的数学方法想了很久，自认为应该是正解了。
看到有多程序贴着，但觉得不少都是骗来的。因为数据特别水。
如果讲得不对，或是不清楚，欢迎大家指出来。QQ841249284
下面就发表一些拙见：
就样例而言：3 60
首先当然要判断 60 mod 3是否为0，若不为0，则立即halt；
k:=60 div 3;//=20
然后把k分解质因数，得到：
2 5
2 1
x，y分别是指两个待求的数
显然这两个2，要不全部都给x，要不都是y.否则如果分别给x,y，那么最大公约数就变成6了，所以就有2种方案
剩下的5，也是这样，可以x，y分别放，又是2中可能。所以答案就是2*2=4。
按照这种思想，就可以将k分解质因数，然后算出k有几个不同的因数，然后答案就为2^k.

DXE—IOU　　　　　　　　　　　　　SYF

没发现m%n!=0.......
还是水过了。。
VijosEx via JudgeDaemon2/13.6.5.0 via libjudge
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 436 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 444 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 444 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 444 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 436 KiB, score = 10
Accepted, time = 0 ms, mem = 444 KiB, score = 50
#include <cstdio>
#include <cmath>

using namespace std;

int n,m;

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

int main(){
scanf("%d%d",&n,&m);
if (m%n) {
printf("0\n");
return 0;
}
int x=m/n;
int ans=0;
for (int i=0;i++<int(sqrt(x));){
if (x%i==0){
if (gcd(x/i,i)==1){
if (x/i!=i) ans+=2; else ans+=1;
}
}
}
printf("%d\n",ans);
return 0;
}

program pxxx;

var x,y,i,o:integer;

begin

readln(x,y);

if y mod x0 then

begin

　　writeln(0);

　　halt;

end;

o:=0;

y:=y div x;

for i:=2 to (y-1) do

if y mod i=0 then

begin

　　inc(o);

　　while y mod i =0 do y:=y div i;

end;

writeln(o*(o-1)+2);

end.

直接ac