莫队算法

#bzoj2038
##600ms
##小Z的袜子
##优化过后的代码，效果强劲……

 $\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$

```c++
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<stack>
#define rez(i,x,y) for(int i=x;i>=y;i--)
#define res(i,x,y) for(int i=x;i<=y;i++)
#define INF 2100000000
#define ll long long
#define clr(x)  memset(x,0,sizeof(x))

using namespace std;
const int maxn = 50000 + 50;
templateinline void readin(T &res){
  static char ch;
  while ((ch = getchar()) < '0' || ch > '9');
  	res = ch - 48;
  while ((ch = getchar()) >= '0' && ch <= '9')
  	res = res * 10 + ch - 48;
}
int n,m,unity,num[maxn],a[maxn],tot,L=1,R=0;;
ll gcd(ll a,ll b){
	return b==0?a:gcd(b,a%b);
}
struct Query{
	int l,r,pos;
	bool operator < (const Query &rhs)const{
		return (l/unity!=rhs.l/unity)?(l/unity<rhs.l/unity):(r<rhs.r);
	}
}node[maxn];
struct Ans{
	ll a,b;
	void reduce(){
		ll dv=gcd(a,b);
		a/=dv,b/=dv;
	}
}ans[maxn];
int main(){
	freopen("in.txt","r",stdin);
	readin(n),readin(m);
	unity=(int)sqrt(n);
	memset(num,0,sizeof(num));
	for(int i=1;i<=n;i++)
		readin(a[i]);
	for(int i=1;i<=m;i++)
		readin(node[i].l),readin(node[i].r),node[i].pos=i;
	sort(node+1,node+1+m);
	for(int i=1;i<=m;i++){
		while(R<node[i].r){
			++R;
			tot+=num[a[R]]*2+1;
			++num[a[R]];
		}
		while(R>node[i].r){
			--num[a[R]];
			tot-=num[a[R]]*2+1;
			--R;
		}
		while(L<node[i].l){
			--num[a[L]];
			tot-=num[a[L]]*2+1;
			++L;
		}
		while(L>node[i].l){
			--L;
			tot+=num[a[L]]*2+1;
			++num[a[L]];
		}
		ans[node[i].pos].a=tot-(R-L+1);
		ans[node[i].pos].b=(ll)(R-L+1)*(R-L);
		ans[node[i].pos].reduce();
	}
	for(int i=1;i<=m;i++)
		//printf("%I64d/%I64d\n",ans[i].a,ans[i].b);
		printf("%lld/%lld\n",ans[i].a,ans[i].b);
	return 0;
}
```