【HDU 1695】GCD(モビウス反転)

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).
直接モビウスで反転すればいいです.∑ i = 1 b ∑ j = 1 d [ g c d ( i , j ) = = k ]\sum_{i=1}^{b}\sum_{j=1}^{d}[gcd(i,j)==k] ∑i=1b​∑j=1d​[gcd(i,j)==k]
k k kを除去するとΣi=1 nΣj=1 m[gcd(i,j)==1]sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)==1] ∑i=1n​∑j=1m​[gcd(i,j)==1]
ここで、n=b/k、m=d/k n=b/k、m=d/k n=b/k、m=d/kである.
f(d)=Σi=1 nΣj=1 m[gcd(i,j)==d]f(d)=sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)==d] f(d)=∑i=1n​∑j=1m​[gcd(i,j)==d].
F(d)=Σi=1 nΣj=1 m[d∣g c d(i,j)]F(d)=sum_{i=1}^{n}\sum_{j=1}^{m}[d|gcd(i,j)] F(d)=∑i=1n​∑j=1m​[d∣gcd(i,j)].
F(d)=Σi∣d f(i)F(d)=sum_{i|d}f(i) F(d)=∑i∣d​f(i)
モビウスの反転:f(d)=Σd∣iμ ( i/d ) ∗ F ( i ) f(d)=\sum_{d|i}\mu(i/d)*F(i) f(d)=∑d∣i​μ(i/d)∗F(i).
従ってf(1)=Σi=1 mi n(n,m)μ ( i ) F ( i ) f(1)=\sum_{i=1}^{min(n,m)}\mu(i)F(i) f(1)=∑i=1min(n,m)​μ(i)F(i)
一方、F(i)=⌊(n/i)⌋(m/i)⌋F(i)=lfloor(n/i)rfloor*lfloor(m/i)rfloor F(i)=⌊(n/i)⌋
最終的には、次のような反転式が得られます.
∑ i = 1 m i n ( n , m ) μ ( i ) ∗ ⌊ ( n/i ) ⌋ ∗ ⌊ ( m/i ) ⌋\sum_{i=1}^{min(n,m)}\mu(i)*\lfloor(n/i)\rfloor*\lfloor(m/i)\rfloor ∑i=1min(n,m)​μ(i)∗⌊(n/i)⌋∗⌊(m/i)⌋
しかし,我々が求めているのは題意とは異なり,題名規定(i,j)=(i,j)=(j,i)=(i,j)=(j,j)=(j,i)であるため,繰返しを計算すると,a=c=1 a=1 a=1 a=c=1となるが,b,d b,dを考慮するだけでよく,繰返しの部分は[1,m i n(b,d)][1,min(b,d)][1,min(b,d)]にしか存在しないことが分かった[1,min(b,d)]では、この段落ではちょうど2回計算して、直接減算すればいいです.最後に(k=0)の場合にも注意します.さもないとREになるかもしれない

#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
",solve(b,d)-solve(min(b,d),min(b,d))/2);
	}
	return 0;
}