GCD HDU - 1695

GCD HDU - 1695

題意：

給出a,b,c,d,k,求出a<=x<=b, c<=y<=d 且gcd(x,y) == k 的（x,y）的對數。

求的是不同數量對的總數

題解：

和這個題一樣P3455 [POI2007]ZAP-Queries，但是本題要求求不同數量對的總數，是以最後的結果要減去重複值

如果是莫比烏斯+分塊做法，對于區間[1,b],[1,d],b<d,重複部分是[1,b]部分，是以減去solve(b,b)/2

如果是容斥做法，對于每個i∈[1,b],求[1,d]中互質的數量，那我們減去i與[1,i]中互質的數量

詳細看代碼

代碼：

莫比烏斯+分塊

#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{
    x= 0;
    char c= getchar();
    bool flag= 0;
    while (c < '0' || c > '9')
        flag|= (c == '-'), c= getchar();
    while (c >= '0' && c <= '9')
        x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();
    if (flag)
        x= -x;
    read(Ar...);
}
template <typename T> inline void write(T x)
{
    if (x < 0) {
        x= ~(x - 1);
        putchar('-');
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef ONLINE_JUDGE
#else
    startTime = clock ();
    freopen("data.in", "r", stdin);
#endif
}
void Time_test()
{
#ifdef ONLINE_JUDGE
#else
    endTime= clock();
    printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn=2e6+9;
int prime[maxn];
int mu[maxn];
int sum[maxn];
int vis[maxn];
int cnt=0;
void get_mu(int N){
	mu[1]=1;
	vis[1]=vis[0]=1;
	for(int i=2;i<=N;i++){
		if(!vis[i]){
			prime[++cnt]=i;
			mu[i]=-1;
		}
		for(int j=1;j<=cnt&&i*prime[j]<=N;j++){
			vis[i*prime[j]]=1;
			if(i%prime[j]==0)break;
			mu[i*prime[j]]=-mu[i];
		}
	}
	for(int i=1;i<=N;i++){
		sum[i]=sum[i-1]+mu[i];
	}
}
ll solve(int a,int b,int k){
	a/=k;
	b/=k;
	int minn=min(a,b);
	ll ans=0;
	for(int l=1,r;l<=minn;l=r+1){
		r=min(a/(a/l),b/(b/l));
		ans+=1ll*(sum[r]-sum[l-1])*(a/l)*(b/l);
	}
	return ans;
}
int main()
{
    //rd_test();
	get_mu(1000000);
	int t;
	read(t);
	int cas=0;
	while(t--){
		int a,b,c,d,k;
		read(a,b,c,d,k);
		if(k==0){
			printf("Case %d: 0\n",++cas);
			continue;
		}
		if(b>d)swap(b,d);
//		printf("%d %d %d %d\n",a,b,c,d); 
		printf("Case %d: %lld\n",++cas,solve(b,d,k)-solve(b,b,k)/2);
	}
    //Time_test();
}




           

容斥

#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{
    x= 0;
    char c= getchar();
    bool flag= 0;
    while (c < '0' || c > '9')
        flag|= (c == '-'), c= getchar();
    while (c >= '0' && c <= '9')
        x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();
    if (flag)
        x= -x;
    read(Ar...);
}
template <typename T> inline void write(T x)
{
    if (x < 0) {
        x= ~(x - 1);
        putchar('-');
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef ONLINE_JUDGE
#else
    startTime = clock ();
    freopen("data.in", "r", stdin);
#endif
}
void Time_test()
{
#ifdef ONLINE_JUDGE
#else
    endTime= clock();
    printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn=3e5+9;
int prime[maxn];
int cnt=0;
void divide(int n){
	cnt=0;
	for(int i=2;i*i<=n;i++){
		if(n%i==0){
			prime[cnt++]=i;
			while(n%i==0)n/=i;
		}
	}
	if(n!=1)prime[cnt++]=n;
} 
int solve(int S){
	int ans=0;
	for(int i=1;i<(1<<cnt);i++){
		int tmp=1;
		int num=0;
		for(int j=0;j<cnt;j++){
			if(i&(1<<j)){
				tmp*=prime[j];
				num++;
			}
		}
		if(num&1)ans+=S/tmp;
		else ans-=S/tmp;
	}
	return S-ans;
}
int main()
{
    //rd_test();
	int t;
	read(t);
	int cas=0;
	while(t--){
		int a,b,c,d,k;
		read(a,b,c,d,k);
		if(k==0){
			printf("Case %d: 0\n",++cas);
			continue;
		}
		if(b>d)swap(b,d);
		b/=k;
		d/=k;
		ll ans=0;
		for(int i=1;i<=b;i++){
			divide(i);
			ans+=solve(d)-solve(i-1);
		}
		printf("Case %d: %lld\n",++cas,ans);
	}
	return 0;
    //Time_test();
}