【HDU 1695】GCD（莫比乌斯反演）

GCD
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16412 Accepted Submission(s): 6314

Problem Description
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).

直接莫比乌斯反演就好了。
${\sum }_{i=1}^b{\sum }_{j=1}^d[gcd(i,j)==k]$

我们把$k$除掉，则有:
${\sum }_{i=1}^n{\sum }_{j=1}^m[gcd(i,j)==1]$

其中$n=b/k，m=d/k$。

我们令$f(d)={\sum }_{i=1}^n{\sum }_{j=1}^m[gcd(i,j)==d]$。

令$F(d)={\sum }_{i=1}^n{\sum }_{j=1}^m[d|gcd(i,j)]$。

则$F(d)={\sum }_{i|d}f(i)$

莫比乌斯反演得:
$f(d)={\sum }_{d|i}\mu (i/d)\ast F(i)$。

因此$f(1)={\sum }_{i=1}^{min(n,m)}\mu (i)F(i)$

而$F(i)=\lfloor (n/i)\rfloor \ast \lfloor (m/i)\rfloor$。

最终我们得到反演公式为：

${\sum }_{i=1}^{min(n,m)}\mu (i)\ast \lfloor (n/i)\rfloor \ast \lfloor (m/i)\rfloor$

但是我们求的和题意不太一样，因为题目规定$(i,j)==(j,i)$，因此我们计算重复了，那么我们怎么去重呢，题目中说$a=c=1$，我们只需要考虑
$b,d$即可，我们发现重复的部分只可能存在于$[1,min(b,d)]$中，在这一段我们恰好计算了两次，直接减去就可以了。
最后还要注意(k==0)的情况。不然可能会RE

#include
using namespace std;
const int maxn = 1e6+10;
const int INF = 0x3f3f3f3f;
typedef long long ll;
ll prime[maxn+100];
bool vis[maxn+100];
int mbs[maxn+100];
ll sum[maxn+100];
void mobious(){
	mbs[1]=1;
	int cnt=0;
	for(int i=2;i<=maxn;i++){
		if(!vis[i]){
			prime[++cnt]=i;
			mbs[i]=-1;
		}
		for(int j=1;j<=cnt&&i*prime[j]<=maxn;j++){
			vis[i*prime[j]]=1;
			if(i%prime[j]==0){
				mbs[i*prime[j]]=0;
				break;
			}
			else mbs[i*prime[j]]=-mbs[i];
		}
	}
	for(int i=1;i<=maxn;i++){
		sum[i]=sum[i-1]+mbs[i];
	}
}
ll k;
ll solve(ll n,ll m){
	n/=k;
	m/=k;
	if(n>m) swap(n,m);
	ll ans=0;
	ll nxt;
	for(int i=1;i<=n;i=nxt+1){
		nxt=min(n/(n/i),m/(m/i));
		ans+=(sum[nxt]-sum[i-1])*(n/i)*(m/i);
	}
	return ans;
}
int main(){
	int cas=1;
	int t;
	mobious();
	scanf("%d",&t);
	ll a,b,c,d;
	while(t--){
		scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&k);
		printf("Case %d: ",cas++);
		if(k==0) puts("0");
		else
		printf("%lld\n",solve(b,d)-solve(min(b,d),min(b,d))/2);
	}
	return 0;
}