題目

思路

設 a,b,c……為每種顏色出現的次數； 答案是：

(a∗(a−1)/2+b∗(b−1)/2+c∗(c−1)/2…)/((R−L+1)∗(R−L)/2)

化簡一下:

(a²+b²+c²+……-(a+b+c+……))/(R-L+1)(R-L)

那麼就是

(a²+b²+c²+……-(R-L+1))/(R-L+1)(R-L)

於是，直接莫隊！

程式碼

#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define ll long long
using namespace std;
const int N=5e4+77,M=5e4+77;
struct ahah
{
	ll L,R,p,f;
}ask[N];
struct haha
{
	ll x,y;
}ans[N];
ll answer,n,m,q,a[N],cnt[N],k;
ll gcd(ll a,ll b)
{
	return !b?a:gcd(b,a%b);
}
bool comp(ahah a,ahah b)
{
	return a.L/k==b.L/k?a.R<b.R:a.L<b.L;
}
void remove(ll pos)
{
	if(--cnt[a[pos]]>0)answer+=(-2)*(cnt[a[pos]]);}
void add(ll pos){ if(++cnt[a[pos]]>1)answer+=2*(cnt[a[pos]]-1); }
void work(ll x,ll y,ll p)
{
	if(!x)ans[p].x=0,ans[p].y=1;
	else 
	{
		ll k=gcd(x,y);
		ans[p].x=x/k;
		ans[p].y=y/k;
	}
}
int main()
{
	scanf("%lld%lld",&n,&q);
	k=sqrt(n);
	for(int i=1; i<=n; i++)scanf("%lld",&a[i]);
	for(int i=1; i<=q; i++)scanf("%lld%lld",&ask[i].L,&ask[i].R),ask[i].p=i;
	sort(ask+1,ask+1+q,comp);
	ll curl=0,curr=0;
	for(int i=1; i<=q; i++)
	{
		ll L=ask[i].L,R=ask[i].R;
		if(L==R)
		{
			ans[ask[i].p].x=0;
			ans[ask[i].p].y=1;
			continue;
		}
		while(curl<L)remove(curl++);
		while(curl>L)add(--curl);
		while(curr<R)add(++curr);
		while(curr>R)remove(curr--);
		work(answer,(R-L+1)*(R-L),ask[i].p);
	}
	for(int i=1; i<=q; i++)printf("%lld/%lld\n",ans[i].x,ans[i].y);
}