今天好颓啊，上下午才打了两个容斥题
容斥定理？不就是小学的韦恩图吗，加加减减的，这有什么大道理？
然后我看到了今天的讲义!和题目！
A - Co-prime
Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.
Input
The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 10 15) and (1 <=N <= 10 9).
Output
For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.
Sample Input
2
1 10 2
3 15 5
Sample Output
Case #1: 5
Case #2: 10
题意：求出a,b间与n不互质的数的个数 好简单的题意
题解：要是直接让你求a,b之间的互质的数，好难啊，那求不互质的数？这可以有，先将n质因数分解，b除以n的质因数就可以得到1~b之间的与n不互质的数的一部分，但你会发现其中有些交集。
比如10中有几个数与6互质，6的质因数为2,3。 10-10/3-10/2=2，，但其中的6这个数被多减了1次所以还要+10/6，这就要用到容斥定理了。
当有偶数种个质因数与n相同，就+，奇数个的就-，b-f[1]+f[2]+....+(-1)^m*f[m].
var
a,b,n,numm,ans,ans1,val,tot:int64;
t,l,i,j:longint;
num:array[0..100005]of int64;
procedure get(x:int64);
var i:longint; k:int64;
begin
k:=n;
for i:=2 to trunc(sqrt(n))+1 do
begin
if k mod i=0 then begin inc(tot); num[tot]:=i; end;
while k mod i=0 do k:=k div i;
end;
if k>1 then begin inc(tot);num[tot]:=k; end;
end;
begin
readln(t);
for l:=1 to t do
begin
fillchar(num,sizeof(num),0); tot:=0;
readln(a,b,n);
get(n);
// for i:=1 to tot do writeln(num[i]);
ans:=b;
for i:=1 to (1<<tot)-1 do
begin
val:=1;numm:=0;
for j:=0 to tot-1 do
if (i and (1<<j))<>0 then
begin
inc(numm);
val:=valnum[j+1];
// writeln(j,' ',i);
end;
if numm mod 2=1 then ans:=ans- b div val else ans:=ans+b div val;
end;
ans1:=a-1;
// writeln(ans);
for i:=1 to (1<<tot)-1 do
begin
val:=1;numm:=0;
for j:=0 to tot-1 do
if ((1<<j) and i)<>0 then
begin
inc(numm);
val:=valnum[j+1];
end;
if numm mod 2=1 then ans1:=ans1- (a-1) div val else ans1:=ans1+(a-1) div val;
end;
writeln('Case #',l,': ',ans-ans1);
end;
end.