这道题弄了很久，好不容易看懂了题解，是用分解质因数+乘法原理，结果写出来的程序还是超时，程序还是太弱了...先贴上来，以后再重新写过
program son;
type
arr=array[1..500] of longint;
var
data,a0,a1,b0,b1:arr;
x0,y0,x1,y1,n,len,ans,k:longint;
procedure first;
var
i,j:longint;
begin
len:=2;
data[1]:=2;
data[2]:=3;
for i:=5 to y1 do
begin
j:=1;
while (j<=len) and (i mod data[j]<>0) do inc(j);
if j=len+1 then
begin
inc(len);
data[len]:=i;
end;
end;
end;
procedure work(var a:arr;k:longint);
var
i:longint;
begin
for i:=1 to len do
begin
if data[i]>k then break;
while (k>0) and (k mod data[i]=0) do
begin
inc(a[i]);
k:=k div data[i];
end;
end;
end;
procedure suan;
var
i,l,r:longint;
begin
for i:=1 to len do
if (a0[i]<>0) or (a1[i]<>0) or (b0[i]<>0) or (b1[i]<>0) then
begin
if (a1[i]>a0[i]) or (b1[i]<b0[i]) then
begin
writeln('0');
{close(input);
close(output);}
halt;
end;
if (a1[i]=a0[i]) or (b1[i]=b0[i]) then
begin
if a1[i]=a0[i] then l:=a1[i];
if b1[i]=b0[i] then r:=b1[i];
end;
if (a1[i]<a0[i]) or (b1[i]>b0[i]) then
begin
if a1[i]<a0[i] then l:=a1[i];
if b1[i]>b0[i] then r:=b1[i];
end;
if (r<l) or (a1[i]<a0[i]) and (b1[i]>b0[i]) and (l<>r) then
begin
writeln('0');
{close(input);
close(output);}
halt;
end;
if r-l>0 then ans:=ans*(r-l);
end;
end;
begin
assign(input,'son.in');
reset(input);
assign(output,'son.out');
rewrite(output);
readln(n);
for k:=1 to n do
begin
fillchar(a0,sizeof(a0),0);
fillchar(a1,sizeof(a1),0);
fillchar(b0,sizeof(b0),0);
fillchar(b1,sizeof(b1),0);
readln(x0,x1,y0,y1);
first;
work(a0,x0);
work(a1,x1);
work(b0,y0);
work(b1,y1);
ans:=1;
suan;
writeln(ans);
end;
close(input);
close(output);
end.
