Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number pairs.
Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.
Yoiu can assume that a = c = 1 in all test cases.
Input
The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.
Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.
Output
For each test case, print the number of choices. Use the format in the example.
Sample Input
2 1 3 1 5 1 1 11014 1 14409 9
Sample Output
Case 1: 9 Case 2: 736427
Hint
For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).
解题思路：
求（1，b）区间和（1，d）区间里面gcd(x, y) = k的数的对数（1<=x<=b , 1<= y <= d）。
先转换成gcd(x/k,y/k)=1的对数，所以区间的范围为(1,b/k)和(1,d/k)，这道题目还要求1-3 和 3-1 这种情况算成一种，因此只需要限制x<y就可以了,
这里使用了容斥定理：
容斥原理有好多写法具体参考博客：https://blog.csdn.net/qq_40707370/article/details/82428221
AC代码：
#include<iostream>
#include<cstdio>
using namespace std;
typedef long long ll;
ll p[10],k;//数组p储存质因子 long long型不同的质因子个数不超过10
void getp(ll n)//分解质因子
{
k=0;
for(int i=2;i*i<=n;i++)
{
if(n%i==0)
p[k++]=i;
while(n%i==0)
n/=i;
}
if(n!=1)
p[k++]=n;
}
ll nod(ll n)//容斥原理
{
ll i,j,top=0,que[1000],sum=0,t;
que[top++]=-1;
for(i=0;i<k;i++)
{
t=top;
for(j=0;j<t;j++)
que[top++]=que[j]*p[i]*-1;
}
for(i=1;i<top;i++)
{
sum+=n/que[i];
}
return n-sum;//返回区间（1，n）中与n是互质的个数
}
int main()
{
ll a,b,c,d,k,t,m,n;
cin>>t;
for(int i=1;i<=t;i++)
{
cin>>a>>b>>c>>d>>k;
if(!k)
{
printf("Case %d: 0\n",i);
continue;
}
b/=k;
d/=k;
if(b==0||d==0)//这一点刚开始没考虑到，wa了，调试了好久好久才找到错误
{
printf("Case %d: 0\n",i);
continue;
}
if(b>d)
swap(b,d);
ll sum=d;//因为1与任何数互质，所以刚开始加d
for(int j=2;j<=b;j++)//在小区间（1，b)从2开始遍历，在区间（1，d)中找与之互质的数
{
getp(j);
sum=sum+nod(d)-nod(j-1); //减去node(j-1)是为了去重（1，2），（2，1）这种情况
}
printf("Case %d: %lld\n",i,sum);
}
return 0;
}